Mixing
Conditional expectation of minimum of exponential random variables
$begingroup$
Let $X_1$ and $X_2$ be independent exponentially distributed random variables with parameter $theta > 0$. I want to compute $mathbb E[X_1 wedge X_2 | X_1]$, where $X_1 wedge X_2 := min(X_1, X_2)$.
I'm really not sure how to do this. I don't want to use any joint distribution formulas (that's a different exercise in this text). Basically all I know about conditional expectations is that $mathbb Eleft[mathbb E[X | Y] mathbb 1_A right] = mathbb E[X mathbb 1_A]$, for any $A in sigma(Y)$. I thought about using this property to calculate $mathbb Eleft[(X_1 wedge X_2) mathbb 1_{{X_1 leq X_2}}| X_1right]$ and $mathbb Eleft[(X_1 wedge X_2) mathbb 1_{{X_1 > X_2}}| X_1right]$ separately, but it's not clear to me that either of these sets are necessarily in $sigma(X_1)$. Any hints?
Edit: I want to avoid using conditional probability over expectations while conditioning over zero-probability events. That's a different section of the book I'm reading out of (Achim Klenke's "Probability Theory: A Comprehensive Course").
Edit 2: I eventually found my own solution, which I've posted as an answer below.
probability probability-theory conditional-expectation conditional-probability exponential-distribution
asked Jan 25 at 4:05
D FordD Ford
665313
$endgroup$
|
show 1 more comment
$begingroup$
Let $X_1$ and $X_2$ be independent exponentially distributed random variables with parameter $theta > 0$. I want to compute $mathbb E[X_1 wedge X_2 | X_1]$, where $X_1 wedge X_2 := min(X_1, X_2)$.
I'm really not sure how to do this. I don't want to use any joint distribution formulas (that's a different exercise in this text). Basically all I know about conditional expectations is that $mathbb Eleft[mathbb E[X | Y] mathbb 1_A right] = mathbb E[X mathbb 1_A]$, for any $A in sigma(Y)$. I thought about using this property to calculate $mathbb Eleft[(X_1 wedge X_2) mathbb 1_{{X_1 leq X_2}}| X_1right]$ and $mathbb Eleft[(X_1 wedge X_2) mathbb 1_{{X_1 > X_2}}| X_1right]$ separately, but it's not clear to me that either of these sets are necessarily in $sigma(X_1)$. Any hints?
Edit: I want to avoid using conditional probability over expectations while conditioning over zero-probability events. That's a different section of the book I'm reading out of (Achim Klenke's "Probability Theory: A Comprehensive Course").
Edit 2: I eventually found my own solution, which I've posted as an answer below.
probability probability-theory conditional-expectation conditional-probability exponential-distribution
asked Jan 25 at 4:05
D FordD Ford
665313
$endgroup$
$begingroup$
You cannot answer this without any independence assumption.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 5:47
$begingroup$
Assuming independence, you might to try to find $mathbb E[k wedge X_2 ]$ for some constant $k ge 0$
$endgroup$
– Henry
Jan 25 at 8:49
$begingroup$
Ah yes, they are independent. I edited it.
$endgroup$
– D Ford
Jan 25 at 16:33
$begingroup$
@Henry I'm not sure how that helps if we assume independence. Is there an independence property you're invoking?
$endgroup$
– D Ford
Jan 25 at 16:34
$begingroup$
With independence, knowing $X_1=k$ tells you nothing new about the distribution of $X_2$, so my hint is helpful. Without independence, my suggestion would not help with the original question
$endgroup$
– Henry
Jan 25 at 16:41
|
show 1 more comment
$begingroup$
Let $X_1$ and $X_2$ be independent exponentially distributed random variables with parameter $theta > 0$. I want to compute $mathbb E[X_1 wedge X_2 | X_1]$, where $X_1 wedge X_2 := min(X_1, X_2)$.
I'm really not sure how to do this. I don't want to use any joint distribution formulas (that's a different exercise in this text). Basically all I know about conditional expectations is that $mathbb Eleft[mathbb E[X | Y] mathbb 1_A right] = mathbb E[X mathbb 1_A]$, for any $A in sigma(Y)$. I thought about using this property to calculate $mathbb Eleft[(X_1 wedge X_2) mathbb 1_{{X_1 leq X_2}}| X_1right]$ and $mathbb Eleft[(X_1 wedge X_2) mathbb 1_{{X_1 > X_2}}| X_1right]$ separately, but it's not clear to me that either of these sets are necessarily in $sigma(X_1)$. Any hints?
Edit: I want to avoid using conditional probability over expectations while conditioning over zero-probability events. That's a different section of the book I'm reading out of (Achim Klenke's "Probability Theory: A Comprehensive Course").
Edit 2: I eventually found my own solution, which I've posted as an answer below.
probability probability-theory conditional-expectation conditional-probability exponential-distribution
asked Jan 25 at 4:05
D FordD Ford
665313
$endgroup$
Let $X_1$ and $X_2$ be independent exponentially distributed random variables with parameter $theta > 0$. I want to compute $mathbb E[X_1 wedge X_2 | X_1]$, where $X_1 wedge X_2 := min(X_1, X_2)$.
I'm really not sure how to do this. I don't want to use any joint distribution formulas (that's a different exercise in this text). Basically all I know about conditional expectations is that $mathbb Eleft[mathbb E[X | Y] mathbb 1_A right] = mathbb E[X mathbb 1_A]$, for any $A in sigma(Y)$. I thought about using this property to calculate $mathbb Eleft[(X_1 wedge X_2) mathbb 1_{{X_1 leq X_2}}| X_1right]$ and $mathbb Eleft[(X_1 wedge X_2) mathbb 1_{{X_1 > X_2}}| X_1right]$ separately, but it's not clear to me that either of these sets are necessarily in $sigma(X_1)$. Any hints?
Edit: I want to avoid using conditional probability over expectations while conditioning over zero-probability events. That's a different section of the book I'm reading out of (Achim Klenke's "Probability Theory: A Comprehensive Course").
Edit 2: I eventually found my own solution, which I've posted as an answer below.
probability probability-theory conditional-expectation conditional-probability exponential-distribution
probability probability-theory conditional-expectation conditional-probability exponential-distribution
asked Jan 25 at 4:05
D FordD Ford
665313
asked Jan 25 at 4:05
D FordD Ford
665313
edited Jan 27 at 21:21
D Ford
asked Jan 25 at 4:05
D FordD Ford
665313
asked Jan 25 at 4:05
D FordD Ford
665313
asked Jan 25 at 4:05
D FordD Ford
665313
665313
$begingroup$
You cannot answer this without any independence assumption.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 5:47
$begingroup$
Assuming independence, you might to try to find $mathbb E[k wedge X_2 ]$ for some constant $k ge 0$
$endgroup$
– Henry
Jan 25 at 8:49
$begingroup$
Ah yes, they are independent. I edited it.
$endgroup$
– D Ford
Jan 25 at 16:33
$begingroup$
@Henry I'm not sure how that helps if we assume independence. Is there an independence property you're invoking?
$endgroup$
– D Ford
Jan 25 at 16:34
$begingroup$
With independence, knowing $X_1=k$ tells you nothing new about the distribution of $X_2$, so my hint is helpful. Without independence, my suggestion would not help with the original question
$endgroup$
– Henry
Jan 25 at 16:41
|
show 1 more comment
$begingroup$
You cannot answer this without any independence assumption.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 5:47
$begingroup$
Assuming independence, you might to try to find $mathbb E[k wedge X_2 ]$ for some constant $k ge 0$
$endgroup$
– Henry
Jan 25 at 8:49
$begingroup$
Ah yes, they are independent. I edited it.
$endgroup$
– D Ford
Jan 25 at 16:33
$begingroup$
@Henry I'm not sure how that helps if we assume independence. Is there an independence property you're invoking?
$endgroup$
– D Ford
Jan 25 at 16:34
$begingroup$
With independence, knowing $X_1=k$ tells you nothing new about the distribution of $X_2$, so my hint is helpful. Without independence, my suggestion would not help with the original question
$endgroup$
– Henry
Jan 25 at 16:41
$begingroup$
You cannot answer this without any independence assumption.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 5:47
$begingroup$
You cannot answer this without any independence assumption.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 5:47
$begingroup$
Assuming independence, you might to try to find $mathbb E[k wedge X_2 ]$ for some constant $k ge 0$
$endgroup$
– Henry
Jan 25 at 8:49
$begingroup$
Assuming independence, you might to try to find $mathbb E[k wedge X_2 ]$ for some constant $k ge 0$
$endgroup$
– Henry
Jan 25 at 8:49
$begingroup$
Ah yes, they are independent. I edited it.
$endgroup$
– D Ford
Jan 25 at 16:33
$begingroup$
Ah yes, they are independent. I edited it.
$endgroup$
– D Ford
Jan 25 at 16:33
$begingroup$
@Henry I'm not sure how that helps if we assume independence. Is there an independence property you're invoking?
$endgroup$
– D Ford
Jan 25 at 16:34
$begingroup$
@Henry I'm not sure how that helps if we assume independence. Is there an independence property you're invoking?
$endgroup$
– D Ford
Jan 25 at 16:34
$begingroup$
With independence, knowing $X_1=k$ tells you nothing new about the distribution of $X_2$, so my hint is helpful. Without independence, my suggestion would not help with the original question
$endgroup$
– Henry
Jan 25 at 16:41
$begingroup$
With independence, knowing $X_1=k$ tells you nothing new about the distribution of $X_2$, so my hint is helpful. Without independence, my suggestion would not help with the original question
$endgroup$
– Henry
Jan 25 at 16:41
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
$$
E[X_1wedge X_2|X_1=s]=int_0^infty (twedge s)hspace{-.5cm}underbrace{theta e^{-theta t}}_{text{conditional pdf}\text{of $X_2$ given $X_1=s$}}hspace{-.5cm},dt=int_0^sttheta e^{-ttheta},dt+int_s^infty stheta e^{-ttheta},dt
$$
answered Jan 25 at 18:31
Mike EarnestMike Earnest
25k22151
$endgroup$
$begingroup$
See my above comment: $mathbb P[X_1 = s] = 0$ for each fixed $s$, and I'd prefer to avoid conditioning on zero-probability events. Plus I'm not sure how to calculate $mathbb E[X_1 wedge X_2 | X_1] $ from knowing $mathbb E[X_1 wedge X_2 | X_1 = s]$ for any given $s$.
$endgroup$
– D Ford
Jan 25 at 20:16
$begingroup$
@DFord The above calculation will give you the correct answer, and then you can prove it is correct by showing it satisfies $E[E[X_1wedge X_2|X_1]1_A]=E[X_1wedge X_2cdot 1_A]$ for all $Ain sigma(X_1)$.
$endgroup$
– Mike Earnest
Jan 25 at 20:20
$begingroup$
Also, $E[X_1wedge X_2|X_1=s]$ works out to be $f(s)$ for some function $s$. Then $E[X_1wedge X_2|X_1]=f(X_1)$. @DFord
$endgroup$
– Mike Earnest
Jan 25 at 20:21
$begingroup$
If $mathbb E[X_1 wedge X_2 | X_1 = s] = f(s)$, wouldn't it follow that $f(X_1) = mathbb E[X_1 wedge X_2 | X_1 = X_1] = mathbb E[X_1 wedge X_2]$?
$endgroup$
– D Ford
Jan 25 at 20:33
1
$begingroup$
@DFord It doesn't quite work like that. If $F(x)=P(Xle x)$ is the cdf of a random variable $X$, do you conclude $F(X)=P(Xle X)=1$? No, $F(X)$ is just another random variable, which takes values between $0$ and $1$. Same kind of thing here.
$endgroup$
– Mike Earnest
Jan 25 at 20:57
|
show 1 more comment
$begingroup$
Here you are an other way to see it (to not condition on zero probability events).
The conditional expectation with respect to a continuous random variable $mathbb{E}[Xmid Y]$ can be defined as (see https://en.wikipedia.org/wiki/Conditional_expectation)
begin{align}
g(y)=lim _{epsilon to 0}mathbb{E} [ X| {omega: |Y(omega)-y |le epsilon} ]
end{align}
Applying this to your case and letting $H_{epsilon,y}={omega: |X_1(omega)-y |le epsilon}$, we get
begin{align}
g(y)
& =lim _{epsilon to 0}mathbb{E} [ X_1 vee X_2 mid H_{epsilon,y} ] \
%
& =lim _{epsilon to 0}frac{mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]}{P(H_{epsilon,y} )} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{mathbb{R}_+^2} (x_1 vee x_2) I_{{|x_1-y |le epsilon}} theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x
}
{int_{y-epsilon}^{y+epsilon} theta e^{-theta z}{rm d }z} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{y-epsilon}^{y+epsilon} int_{mathbb{R}_+} (x_1 vee x_2) theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x_2{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
%
left(
int_{0}^{x_1} x_2theta e^{-theta x_2} {rm d }x_2
+
x_1 int_{x_1}^{infty} theta e^{-theta x_2} {rm d }x_2
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
e^{-2theta x_1}
%
left(
e^{theta x_1} - theta x_1 -1
+
theta x_1
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac{1}{theta }
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
{rm d }x_1
-
int_{y-epsilon}^{y+epsilon}
theta e^{-2theta x_1}
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =
frac{1}{theta }-
frac{1}{2theta }
lim _{epsilon to 0}
frac
{
e^{-2theta(y-epsilon)} -e^{-2theta(y+epsilon)}
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
% & =
% frac{1}{theta }-
% frac{e^{-theta y}}{2theta }
% lim _{epsilon to 0}
% frac
% {
% e^{2theta epsilon} - 1 - (e^{-2theta epsilon} -1)
% }
% {
% e^{ theta epsilon} -1 - (e^{- theta epsilon}-1)
% } \
& =
frac{1}{theta }-
frac{e^{-theta y}}{2theta }
times
frac
{
4 theta
}
{
2 theta
} \
& =
frac{1-e^{-theta y}}{theta }
end{align}
Hope it helps.
answered Jan 26 at 21:47
user52227user52227
1,038512
$endgroup$
$begingroup$
Yes, that's the answer I got. Though I ended up finding it using image measures and densities instead of explicit limits, although I like your solution. Thanks!
$endgroup$
– D Ford
Jan 27 at 20:03
add a comment |
$begingroup$
We're looking for a $sigma(X_1)$-measurable function $mathbb Eleft[X_1 wedge X_2 | X_1right] : Omega to mathbb R$ for which for every $A in sigma(X_1)$,
$$
mathbb Eleft[left(X_1 wedge X_2right) mathbb 1_Aright] = mathbb Eleft[mathbb Eleft[ X_1 wedge X_2 | X_1 right] mathbb 1_A right].
$$
Let $f_theta(x) = theta e^{-theta x}$ be the probability density of $X_1$ and of $X_2$, and let $lambda_n$ denote the $n$-dimensional Lebesgue measure. Then $f_theta , dlambda_1 = dleft(mathbb P circ X_1^{-1}right)$. Furthermore, by independence of $X_1$ and $X_2$, the joint probability density is $tilde f(x,y) = theta^2 e^{-theta(x+y)}$, and $dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) = tilde f , dlambda_2$, where $X_1 times X_2 : Omega to mathbb R^2$ is $omega mapsto left(X_1(omega), X_2(omega)right)$. Let $A in sigma(X_1)$. Then,
begin{align*}
mathbb Eleft[left(X_1 wedge X_2 right)mathbb 1_A right] &= int_A X_1 wedge X_2 , dmathbb P = int_{X_1(A) times [0,infty)} x wedge y , dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) \
&= int_{X_1(A)} int_0^infty left(x wedge yright) theta^2 e^{-theta(x+y)} , dy , dx\ &= int_{X_1(A)} int_0^x theta^2 ye^{-theta(x+y)} , dy , dx + int_{X_1(A)} int_x^infty theta^2 x e^{-theta(x+y)} , dy , dx \
&= int_{X_1(A)} left(-theta x e^{-2theta x} - e^{-2theta x} + e^{-theta x}right) , dx + int_{X_1(A)} theta x e^{-2theta x},dx \
&= int_{X_1(A)} left(e^{-theta x} - e^{-2theta x}right),dx = int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right)theta e^{-theta x},dx \
&= int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right) , dleft(mathbb P circ X_1^{-1}right)(x) = int_A frac 1 theta left(1-e^{-theta X_1}right) , dmathbb P \
&= mathbb Eleft[frac 1 theta left(1-e^{-theta X_1}right) mathbb 1_A right].
end{align*}
From this it follows that $boxed{mathbb Eleft[X_1 wedge X_2 | X_1right] = frac 1 theta left(1-e^{-theta X_1}right).}$
answered Jan 27 at 20:02
D FordD Ford
665313
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
$$
E[X_1wedge X_2|X_1=s]=int_0^infty (twedge s)hspace{-.5cm}underbrace{theta e^{-theta t}}_{text{conditional pdf}\text{of $X_2$ given $X_1=s$}}hspace{-.5cm},dt=int_0^sttheta e^{-ttheta},dt+int_s^infty stheta e^{-ttheta},dt
$$
answered Jan 25 at 18:31
Mike EarnestMike Earnest
25k22151
$endgroup$
$begingroup$
See my above comment: $mathbb P[X_1 = s] = 0$ for each fixed $s$, and I'd prefer to avoid conditioning on zero-probability events. Plus I'm not sure how to calculate $mathbb E[X_1 wedge X_2 | X_1] $ from knowing $mathbb E[X_1 wedge X_2 | X_1 = s]$ for any given $s$.
$endgroup$
– D Ford
Jan 25 at 20:16
$begingroup$
@DFord The above calculation will give you the correct answer, and then you can prove it is correct by showing it satisfies $E[E[X_1wedge X_2|X_1]1_A]=E[X_1wedge X_2cdot 1_A]$ for all $Ain sigma(X_1)$.
$endgroup$
– Mike Earnest
Jan 25 at 20:20
$begingroup$
Also, $E[X_1wedge X_2|X_1=s]$ works out to be $f(s)$ for some function $s$. Then $E[X_1wedge X_2|X_1]=f(X_1)$. @DFord
$endgroup$
– Mike Earnest
Jan 25 at 20:21
$begingroup$
If $mathbb E[X_1 wedge X_2 | X_1 = s] = f(s)$, wouldn't it follow that $f(X_1) = mathbb E[X_1 wedge X_2 | X_1 = X_1] = mathbb E[X_1 wedge X_2]$?
$endgroup$
– D Ford
Jan 25 at 20:33
1
$begingroup$
@DFord It doesn't quite work like that. If $F(x)=P(Xle x)$ is the cdf of a random variable $X$, do you conclude $F(X)=P(Xle X)=1$? No, $F(X)$ is just another random variable, which takes values between $0$ and $1$. Same kind of thing here.
$endgroup$
– Mike Earnest
Jan 25 at 20:57
|
show 1 more comment
$begingroup$
$$
E[X_1wedge X_2|X_1=s]=int_0^infty (twedge s)hspace{-.5cm}underbrace{theta e^{-theta t}}_{text{conditional pdf}\text{of $X_2$ given $X_1=s$}}hspace{-.5cm},dt=int_0^sttheta e^{-ttheta},dt+int_s^infty stheta e^{-ttheta},dt
$$
answered Jan 25 at 18:31
Mike EarnestMike Earnest
25k22151
$endgroup$
$begingroup$
See my above comment: $mathbb P[X_1 = s] = 0$ for each fixed $s$, and I'd prefer to avoid conditioning on zero-probability events. Plus I'm not sure how to calculate $mathbb E[X_1 wedge X_2 | X_1] $ from knowing $mathbb E[X_1 wedge X_2 | X_1 = s]$ for any given $s$.
$endgroup$
– D Ford
Jan 25 at 20:16
$begingroup$
@DFord The above calculation will give you the correct answer, and then you can prove it is correct by showing it satisfies $E[E[X_1wedge X_2|X_1]1_A]=E[X_1wedge X_2cdot 1_A]$ for all $Ain sigma(X_1)$.
$endgroup$
– Mike Earnest
Jan 25 at 20:20
$begingroup$
Also, $E[X_1wedge X_2|X_1=s]$ works out to be $f(s)$ for some function $s$. Then $E[X_1wedge X_2|X_1]=f(X_1)$. @DFord
$endgroup$
– Mike Earnest
Jan 25 at 20:21
$begingroup$
If $mathbb E[X_1 wedge X_2 | X_1 = s] = f(s)$, wouldn't it follow that $f(X_1) = mathbb E[X_1 wedge X_2 | X_1 = X_1] = mathbb E[X_1 wedge X_2]$?
$endgroup$
– D Ford
Jan 25 at 20:33
1
$begingroup$
@DFord It doesn't quite work like that. If $F(x)=P(Xle x)$ is the cdf of a random variable $X$, do you conclude $F(X)=P(Xle X)=1$? No, $F(X)$ is just another random variable, which takes values between $0$ and $1$. Same kind of thing here.
$endgroup$
– Mike Earnest
Jan 25 at 20:57
|
show 1 more comment
$begingroup$
$$
E[X_1wedge X_2|X_1=s]=int_0^infty (twedge s)hspace{-.5cm}underbrace{theta e^{-theta t}}_{text{conditional pdf}\text{of $X_2$ given $X_1=s$}}hspace{-.5cm},dt=int_0^sttheta e^{-ttheta},dt+int_s^infty stheta e^{-ttheta},dt
$$
answered Jan 25 at 18:31
Mike EarnestMike Earnest
25k22151
$endgroup$
$$
E[X_1wedge X_2|X_1=s]=int_0^infty (twedge s)hspace{-.5cm}underbrace{theta e^{-theta t}}_{text{conditional pdf}\text{of $X_2$ given $X_1=s$}}hspace{-.5cm},dt=int_0^sttheta e^{-ttheta},dt+int_s^infty stheta e^{-ttheta},dt
$$
answered Jan 25 at 18:31
Mike EarnestMike Earnest
25k22151
answered Jan 25 at 18:31
Mike EarnestMike Earnest
25k22151
answered Jan 25 at 18:31
Mike EarnestMike Earnest
25k22151
answered Jan 25 at 18:31
Mike EarnestMike Earnest
25k22151
25k22151
$begingroup$
See my above comment: $mathbb P[X_1 = s] = 0$ for each fixed $s$, and I'd prefer to avoid conditioning on zero-probability events. Plus I'm not sure how to calculate $mathbb E[X_1 wedge X_2 | X_1] $ from knowing $mathbb E[X_1 wedge X_2 | X_1 = s]$ for any given $s$.
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– D Ford
Jan 25 at 20:16
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@DFord The above calculation will give you the correct answer, and then you can prove it is correct by showing it satisfies $E[E[X_1wedge X_2|X_1]1_A]=E[X_1wedge X_2cdot 1_A]$ for all $Ain sigma(X_1)$.
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– Mike Earnest
Jan 25 at 20:20
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Also, $E[X_1wedge X_2|X_1=s]$ works out to be $f(s)$ for some function $s$. Then $E[X_1wedge X_2|X_1]=f(X_1)$. @DFord
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– Mike Earnest
Jan 25 at 20:21
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If $mathbb E[X_1 wedge X_2 | X_1 = s] = f(s)$, wouldn't it follow that $f(X_1) = mathbb E[X_1 wedge X_2 | X_1 = X_1] = mathbb E[X_1 wedge X_2]$?
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– D Ford
Jan 25 at 20:33
1
$begingroup$
@DFord It doesn't quite work like that. If $F(x)=P(Xle x)$ is the cdf of a random variable $X$, do you conclude $F(X)=P(Xle X)=1$? No, $F(X)$ is just another random variable, which takes values between $0$ and $1$. Same kind of thing here.
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– Mike Earnest
Jan 25 at 20:57
|
show 1 more comment
$begingroup$
See my above comment: $mathbb P[X_1 = s] = 0$ for each fixed $s$, and I'd prefer to avoid conditioning on zero-probability events. Plus I'm not sure how to calculate $mathbb E[X_1 wedge X_2 | X_1] $ from knowing $mathbb E[X_1 wedge X_2 | X_1 = s]$ for any given $s$.
$endgroup$
– D Ford
Jan 25 at 20:16
$begingroup$
@DFord The above calculation will give you the correct answer, and then you can prove it is correct by showing it satisfies $E[E[X_1wedge X_2|X_1]1_A]=E[X_1wedge X_2cdot 1_A]$ for all $Ain sigma(X_1)$.
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– Mike Earnest
Jan 25 at 20:20
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Also, $E[X_1wedge X_2|X_1=s]$ works out to be $f(s)$ for some function $s$. Then $E[X_1wedge X_2|X_1]=f(X_1)$. @DFord
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– Mike Earnest
Jan 25 at 20:21
$begingroup$
If $mathbb E[X_1 wedge X_2 | X_1 = s] = f(s)$, wouldn't it follow that $f(X_1) = mathbb E[X_1 wedge X_2 | X_1 = X_1] = mathbb E[X_1 wedge X_2]$?
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– D Ford
Jan 25 at 20:33
1
$begingroup$
@DFord It doesn't quite work like that. If $F(x)=P(Xle x)$ is the cdf of a random variable $X$, do you conclude $F(X)=P(Xle X)=1$? No, $F(X)$ is just another random variable, which takes values between $0$ and $1$. Same kind of thing here.
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– Mike Earnest
Jan 25 at 20:57
$begingroup$
See my above comment: $mathbb P[X_1 = s] = 0$ for each fixed $s$, and I'd prefer to avoid conditioning on zero-probability events. Plus I'm not sure how to calculate $mathbb E[X_1 wedge X_2 | X_1] $ from knowing $mathbb E[X_1 wedge X_2 | X_1 = s]$ for any given $s$.
$endgroup$
– D Ford
Jan 25 at 20:16
$begingroup$
See my above comment: $mathbb P[X_1 = s] = 0$ for each fixed $s$, and I'd prefer to avoid conditioning on zero-probability events. Plus I'm not sure how to calculate $mathbb E[X_1 wedge X_2 | X_1] $ from knowing $mathbb E[X_1 wedge X_2 | X_1 = s]$ for any given $s$.
$endgroup$
– D Ford
Jan 25 at 20:16
$begingroup$
@DFord The above calculation will give you the correct answer, and then you can prove it is correct by showing it satisfies $E[E[X_1wedge X_2|X_1]1_A]=E[X_1wedge X_2cdot 1_A]$ for all $Ain sigma(X_1)$.
$endgroup$
– Mike Earnest
Jan 25 at 20:20
$begingroup$
@DFord The above calculation will give you the correct answer, and then you can prove it is correct by showing it satisfies $E[E[X_1wedge X_2|X_1]1_A]=E[X_1wedge X_2cdot 1_A]$ for all $Ain sigma(X_1)$.
$endgroup$
– Mike Earnest
Jan 25 at 20:20
$begingroup$
Also, $E[X_1wedge X_2|X_1=s]$ works out to be $f(s)$ for some function $s$. Then $E[X_1wedge X_2|X_1]=f(X_1)$. @DFord
$endgroup$
– Mike Earnest
Jan 25 at 20:21
$begingroup$
Also, $E[X_1wedge X_2|X_1=s]$ works out to be $f(s)$ for some function $s$. Then $E[X_1wedge X_2|X_1]=f(X_1)$. @DFord
$endgroup$
– Mike Earnest
Jan 25 at 20:21
$begingroup$
If $mathbb E[X_1 wedge X_2 | X_1 = s] = f(s)$, wouldn't it follow that $f(X_1) = mathbb E[X_1 wedge X_2 | X_1 = X_1] = mathbb E[X_1 wedge X_2]$?
$endgroup$
– D Ford
Jan 25 at 20:33
$begingroup$
If $mathbb E[X_1 wedge X_2 | X_1 = s] = f(s)$, wouldn't it follow that $f(X_1) = mathbb E[X_1 wedge X_2 | X_1 = X_1] = mathbb E[X_1 wedge X_2]$?
$endgroup$
– D Ford
Jan 25 at 20:33
1
1
$begingroup$
@DFord It doesn't quite work like that. If $F(x)=P(Xle x)$ is the cdf of a random variable $X$, do you conclude $F(X)=P(Xle X)=1$? No, $F(X)$ is just another random variable, which takes values between $0$ and $1$. Same kind of thing here.
$endgroup$
– Mike Earnest
Jan 25 at 20:57
$begingroup$
@DFord It doesn't quite work like that. If $F(x)=P(Xle x)$ is the cdf of a random variable $X$, do you conclude $F(X)=P(Xle X)=1$? No, $F(X)$ is just another random variable, which takes values between $0$ and $1$. Same kind of thing here.
$endgroup$
– Mike Earnest
Jan 25 at 20:57
|
show 1 more comment
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Here you are an other way to see it (to not condition on zero probability events).
The conditional expectation with respect to a continuous random variable $mathbb{E}[Xmid Y]$ can be defined as (see https://en.wikipedia.org/wiki/Conditional_expectation)
begin{align}
g(y)=lim _{epsilon to 0}mathbb{E} [ X| {omega: |Y(omega)-y |le epsilon} ]
end{align}
Applying this to your case and letting $H_{epsilon,y}={omega: |X_1(omega)-y |le epsilon}$, we get
begin{align}
g(y)
& =lim _{epsilon to 0}mathbb{E} [ X_1 vee X_2 mid H_{epsilon,y} ] \
%
& =lim _{epsilon to 0}frac{mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]}{P(H_{epsilon,y} )} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{mathbb{R}_+^2} (x_1 vee x_2) I_{{|x_1-y |le epsilon}} theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x
}
{int_{y-epsilon}^{y+epsilon} theta e^{-theta z}{rm d }z} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{y-epsilon}^{y+epsilon} int_{mathbb{R}_+} (x_1 vee x_2) theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x_2{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
%
left(
int_{0}^{x_1} x_2theta e^{-theta x_2} {rm d }x_2
+
x_1 int_{x_1}^{infty} theta e^{-theta x_2} {rm d }x_2
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
e^{-2theta x_1}
%
left(
e^{theta x_1} - theta x_1 -1
+
theta x_1
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac{1}{theta }
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
{rm d }x_1
-
int_{y-epsilon}^{y+epsilon}
theta e^{-2theta x_1}
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =
frac{1}{theta }-
frac{1}{2theta }
lim _{epsilon to 0}
frac
{
e^{-2theta(y-epsilon)} -e^{-2theta(y+epsilon)}
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
% & =
% frac{1}{theta }-
% frac{e^{-theta y}}{2theta }
% lim _{epsilon to 0}
% frac
% {
% e^{2theta epsilon} - 1 - (e^{-2theta epsilon} -1)
% }
% {
% e^{ theta epsilon} -1 - (e^{- theta epsilon}-1)
% } \
& =
frac{1}{theta }-
frac{e^{-theta y}}{2theta }
times
frac
{
4 theta
}
{
2 theta
} \
& =
frac{1-e^{-theta y}}{theta }
end{align}
Hope it helps.
answered Jan 26 at 21:47
user52227user52227
1,038512
$endgroup$
$begingroup$
Yes, that's the answer I got. Though I ended up finding it using image measures and densities instead of explicit limits, although I like your solution. Thanks!
$endgroup$
– D Ford
Jan 27 at 20:03
add a comment |
$begingroup$
Here you are an other way to see it (to not condition on zero probability events).
The conditional expectation with respect to a continuous random variable $mathbb{E}[Xmid Y]$ can be defined as (see https://en.wikipedia.org/wiki/Conditional_expectation)
begin{align}
g(y)=lim _{epsilon to 0}mathbb{E} [ X| {omega: |Y(omega)-y |le epsilon} ]
end{align}
Applying this to your case and letting $H_{epsilon,y}={omega: |X_1(omega)-y |le epsilon}$, we get
begin{align}
g(y)
& =lim _{epsilon to 0}mathbb{E} [ X_1 vee X_2 mid H_{epsilon,y} ] \
%
& =lim _{epsilon to 0}frac{mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]}{P(H_{epsilon,y} )} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{mathbb{R}_+^2} (x_1 vee x_2) I_{{|x_1-y |le epsilon}} theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x
}
{int_{y-epsilon}^{y+epsilon} theta e^{-theta z}{rm d }z} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{y-epsilon}^{y+epsilon} int_{mathbb{R}_+} (x_1 vee x_2) theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x_2{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
%
left(
int_{0}^{x_1} x_2theta e^{-theta x_2} {rm d }x_2
+
x_1 int_{x_1}^{infty} theta e^{-theta x_2} {rm d }x_2
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
e^{-2theta x_1}
%
left(
e^{theta x_1} - theta x_1 -1
+
theta x_1
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac{1}{theta }
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
{rm d }x_1
-
int_{y-epsilon}^{y+epsilon}
theta e^{-2theta x_1}
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =
frac{1}{theta }-
frac{1}{2theta }
lim _{epsilon to 0}
frac
{
e^{-2theta(y-epsilon)} -e^{-2theta(y+epsilon)}
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
% & =
% frac{1}{theta }-
% frac{e^{-theta y}}{2theta }
% lim _{epsilon to 0}
% frac
% {
% e^{2theta epsilon} - 1 - (e^{-2theta epsilon} -1)
% }
% {
% e^{ theta epsilon} -1 - (e^{- theta epsilon}-1)
% } \
& =
frac{1}{theta }-
frac{e^{-theta y}}{2theta }
times
frac
{
4 theta
}
{
2 theta
} \
& =
frac{1-e^{-theta y}}{theta }
end{align}
Hope it helps.
answered Jan 26 at 21:47
user52227user52227
1,038512
$endgroup$
$begingroup$
Yes, that's the answer I got. Though I ended up finding it using image measures and densities instead of explicit limits, although I like your solution. Thanks!
$endgroup$
– D Ford
Jan 27 at 20:03
add a comment |
$begingroup$
Here you are an other way to see it (to not condition on zero probability events).
The conditional expectation with respect to a continuous random variable $mathbb{E}[Xmid Y]$ can be defined as (see https://en.wikipedia.org/wiki/Conditional_expectation)
begin{align}
g(y)=lim _{epsilon to 0}mathbb{E} [ X| {omega: |Y(omega)-y |le epsilon} ]
end{align}
Applying this to your case and letting $H_{epsilon,y}={omega: |X_1(omega)-y |le epsilon}$, we get
begin{align}
g(y)
& =lim _{epsilon to 0}mathbb{E} [ X_1 vee X_2 mid H_{epsilon,y} ] \
%
& =lim _{epsilon to 0}frac{mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]}{P(H_{epsilon,y} )} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{mathbb{R}_+^2} (x_1 vee x_2) I_{{|x_1-y |le epsilon}} theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x
}
{int_{y-epsilon}^{y+epsilon} theta e^{-theta z}{rm d }z} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{y-epsilon}^{y+epsilon} int_{mathbb{R}_+} (x_1 vee x_2) theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x_2{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
%
left(
int_{0}^{x_1} x_2theta e^{-theta x_2} {rm d }x_2
+
x_1 int_{x_1}^{infty} theta e^{-theta x_2} {rm d }x_2
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
e^{-2theta x_1}
%
left(
e^{theta x_1} - theta x_1 -1
+
theta x_1
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac{1}{theta }
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
{rm d }x_1
-
int_{y-epsilon}^{y+epsilon}
theta e^{-2theta x_1}
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =
frac{1}{theta }-
frac{1}{2theta }
lim _{epsilon to 0}
frac
{
e^{-2theta(y-epsilon)} -e^{-2theta(y+epsilon)}
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
% & =
% frac{1}{theta }-
% frac{e^{-theta y}}{2theta }
% lim _{epsilon to 0}
% frac
% {
% e^{2theta epsilon} - 1 - (e^{-2theta epsilon} -1)
% }
% {
% e^{ theta epsilon} -1 - (e^{- theta epsilon}-1)
% } \
& =
frac{1}{theta }-
frac{e^{-theta y}}{2theta }
times
frac
{
4 theta
}
{
2 theta
} \
& =
frac{1-e^{-theta y}}{theta }
end{align}
Hope it helps.
answered Jan 26 at 21:47
user52227user52227
1,038512
$endgroup$
Here you are an other way to see it (to not condition on zero probability events).
The conditional expectation with respect to a continuous random variable $mathbb{E}[Xmid Y]$ can be defined as (see https://en.wikipedia.org/wiki/Conditional_expectation)
begin{align}
g(y)=lim _{epsilon to 0}mathbb{E} [ X| {omega: |Y(omega)-y |le epsilon} ]
end{align}
Applying this to your case and letting $H_{epsilon,y}={omega: |X_1(omega)-y |le epsilon}$, we get
begin{align}
g(y)
& =lim _{epsilon to 0}mathbb{E} [ X_1 vee X_2 mid H_{epsilon,y} ] \
%
& =lim _{epsilon to 0}frac{mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]}{P(H_{epsilon,y} )} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{mathbb{R}_+^2} (x_1 vee x_2) I_{{|x_1-y |le epsilon}} theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x
}
{int_{y-epsilon}^{y+epsilon} theta e^{-theta z}{rm d }z} \
%
& =lim _{epsilon to 0}
frac
{
% mathbb{E} [ (X_1 vee X_2) I_{H_{epsilon,y}} ]
int_{y-epsilon}^{y+epsilon} int_{mathbb{R}_+} (x_1 vee x_2) theta e^{-theta x_1} theta e^{-theta x_2} {rm d }x_2{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
%
left(
int_{0}^{x_1} x_2theta e^{-theta x_2} {rm d }x_2
+
x_1 int_{x_1}^{infty} theta e^{-theta x_2} {rm d }x_2
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac
{
int_{y-epsilon}^{y+epsilon}
e^{-2theta x_1}
%
left(
e^{theta x_1} - theta x_1 -1
+
theta x_1
right)
%
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =lim _{epsilon to 0}
frac{1}{theta }
frac
{
int_{y-epsilon}^{y+epsilon}
theta e^{-theta x_1}
{rm d }x_1
-
int_{y-epsilon}^{y+epsilon}
theta e^{-2theta x_1}
{rm d }x_1
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
& =
frac{1}{theta }-
frac{1}{2theta }
lim _{epsilon to 0}
frac
{
e^{-2theta(y-epsilon)} -e^{-2theta(y+epsilon)}
}
{
e^{-theta(y-epsilon)} -e^{-theta(y+epsilon)}
} \
%
% & =
% frac{1}{theta }-
% frac{e^{-theta y}}{2theta }
% lim _{epsilon to 0}
% frac
% {
% e^{2theta epsilon} - 1 - (e^{-2theta epsilon} -1)
% }
% {
% e^{ theta epsilon} -1 - (e^{- theta epsilon}-1)
% } \
& =
frac{1}{theta }-
frac{e^{-theta y}}{2theta }
times
frac
{
4 theta
}
{
2 theta
} \
& =
frac{1-e^{-theta y}}{theta }
end{align}
Hope it helps.
answered Jan 26 at 21:47
user52227user52227
1,038512
edited Jan 27 at 13:20
answered Jan 26 at 21:47
user52227user52227
1,038512
answered Jan 26 at 21:47
user52227user52227
1,038512
answered Jan 26 at 21:47
user52227user52227
1,038512
1,038512
$begingroup$
Yes, that's the answer I got. Though I ended up finding it using image measures and densities instead of explicit limits, although I like your solution. Thanks!
$endgroup$
– D Ford
Jan 27 at 20:03
add a comment |
$begingroup$
Yes, that's the answer I got. Though I ended up finding it using image measures and densities instead of explicit limits, although I like your solution. Thanks!
$endgroup$
– D Ford
Jan 27 at 20:03
$begingroup$
Yes, that's the answer I got. Though I ended up finding it using image measures and densities instead of explicit limits, although I like your solution. Thanks!
$endgroup$
– D Ford
Jan 27 at 20:03
$begingroup$
Yes, that's the answer I got. Though I ended up finding it using image measures and densities instead of explicit limits, although I like your solution. Thanks!
$endgroup$
– D Ford
Jan 27 at 20:03
add a comment |
$begingroup$
We're looking for a $sigma(X_1)$-measurable function $mathbb Eleft[X_1 wedge X_2 | X_1right] : Omega to mathbb R$ for which for every $A in sigma(X_1)$,
$$
mathbb Eleft[left(X_1 wedge X_2right) mathbb 1_Aright] = mathbb Eleft[mathbb Eleft[ X_1 wedge X_2 | X_1 right] mathbb 1_A right].
$$
Let $f_theta(x) = theta e^{-theta x}$ be the probability density of $X_1$ and of $X_2$, and let $lambda_n$ denote the $n$-dimensional Lebesgue measure. Then $f_theta , dlambda_1 = dleft(mathbb P circ X_1^{-1}right)$. Furthermore, by independence of $X_1$ and $X_2$, the joint probability density is $tilde f(x,y) = theta^2 e^{-theta(x+y)}$, and $dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) = tilde f , dlambda_2$, where $X_1 times X_2 : Omega to mathbb R^2$ is $omega mapsto left(X_1(omega), X_2(omega)right)$. Let $A in sigma(X_1)$. Then,
begin{align*}
mathbb Eleft[left(X_1 wedge X_2 right)mathbb 1_A right] &= int_A X_1 wedge X_2 , dmathbb P = int_{X_1(A) times [0,infty)} x wedge y , dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) \
&= int_{X_1(A)} int_0^infty left(x wedge yright) theta^2 e^{-theta(x+y)} , dy , dx\ &= int_{X_1(A)} int_0^x theta^2 ye^{-theta(x+y)} , dy , dx + int_{X_1(A)} int_x^infty theta^2 x e^{-theta(x+y)} , dy , dx \
&= int_{X_1(A)} left(-theta x e^{-2theta x} - e^{-2theta x} + e^{-theta x}right) , dx + int_{X_1(A)} theta x e^{-2theta x},dx \
&= int_{X_1(A)} left(e^{-theta x} - e^{-2theta x}right),dx = int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right)theta e^{-theta x},dx \
&= int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right) , dleft(mathbb P circ X_1^{-1}right)(x) = int_A frac 1 theta left(1-e^{-theta X_1}right) , dmathbb P \
&= mathbb Eleft[frac 1 theta left(1-e^{-theta X_1}right) mathbb 1_A right].
end{align*}
From this it follows that $boxed{mathbb Eleft[X_1 wedge X_2 | X_1right] = frac 1 theta left(1-e^{-theta X_1}right).}$
answered Jan 27 at 20:02
D FordD Ford
665313
$endgroup$
add a comment |
$begingroup$
We're looking for a $sigma(X_1)$-measurable function $mathbb Eleft[X_1 wedge X_2 | X_1right] : Omega to mathbb R$ for which for every $A in sigma(X_1)$,
$$
mathbb Eleft[left(X_1 wedge X_2right) mathbb 1_Aright] = mathbb Eleft[mathbb Eleft[ X_1 wedge X_2 | X_1 right] mathbb 1_A right].
$$
Let $f_theta(x) = theta e^{-theta x}$ be the probability density of $X_1$ and of $X_2$, and let $lambda_n$ denote the $n$-dimensional Lebesgue measure. Then $f_theta , dlambda_1 = dleft(mathbb P circ X_1^{-1}right)$. Furthermore, by independence of $X_1$ and $X_2$, the joint probability density is $tilde f(x,y) = theta^2 e^{-theta(x+y)}$, and $dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) = tilde f , dlambda_2$, where $X_1 times X_2 : Omega to mathbb R^2$ is $omega mapsto left(X_1(omega), X_2(omega)right)$. Let $A in sigma(X_1)$. Then,
begin{align*}
mathbb Eleft[left(X_1 wedge X_2 right)mathbb 1_A right] &= int_A X_1 wedge X_2 , dmathbb P = int_{X_1(A) times [0,infty)} x wedge y , dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) \
&= int_{X_1(A)} int_0^infty left(x wedge yright) theta^2 e^{-theta(x+y)} , dy , dx\ &= int_{X_1(A)} int_0^x theta^2 ye^{-theta(x+y)} , dy , dx + int_{X_1(A)} int_x^infty theta^2 x e^{-theta(x+y)} , dy , dx \
&= int_{X_1(A)} left(-theta x e^{-2theta x} - e^{-2theta x} + e^{-theta x}right) , dx + int_{X_1(A)} theta x e^{-2theta x},dx \
&= int_{X_1(A)} left(e^{-theta x} - e^{-2theta x}right),dx = int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right)theta e^{-theta x},dx \
&= int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right) , dleft(mathbb P circ X_1^{-1}right)(x) = int_A frac 1 theta left(1-e^{-theta X_1}right) , dmathbb P \
&= mathbb Eleft[frac 1 theta left(1-e^{-theta X_1}right) mathbb 1_A right].
end{align*}
From this it follows that $boxed{mathbb Eleft[X_1 wedge X_2 | X_1right] = frac 1 theta left(1-e^{-theta X_1}right).}$
answered Jan 27 at 20:02
D FordD Ford
665313
$endgroup$
add a comment |
$begingroup$
We're looking for a $sigma(X_1)$-measurable function $mathbb Eleft[X_1 wedge X_2 | X_1right] : Omega to mathbb R$ for which for every $A in sigma(X_1)$,
$$
mathbb Eleft[left(X_1 wedge X_2right) mathbb 1_Aright] = mathbb Eleft[mathbb Eleft[ X_1 wedge X_2 | X_1 right] mathbb 1_A right].
$$
Let $f_theta(x) = theta e^{-theta x}$ be the probability density of $X_1$ and of $X_2$, and let $lambda_n$ denote the $n$-dimensional Lebesgue measure. Then $f_theta , dlambda_1 = dleft(mathbb P circ X_1^{-1}right)$. Furthermore, by independence of $X_1$ and $X_2$, the joint probability density is $tilde f(x,y) = theta^2 e^{-theta(x+y)}$, and $dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) = tilde f , dlambda_2$, where $X_1 times X_2 : Omega to mathbb R^2$ is $omega mapsto left(X_1(omega), X_2(omega)right)$. Let $A in sigma(X_1)$. Then,
begin{align*}
mathbb Eleft[left(X_1 wedge X_2 right)mathbb 1_A right] &= int_A X_1 wedge X_2 , dmathbb P = int_{X_1(A) times [0,infty)} x wedge y , dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) \
&= int_{X_1(A)} int_0^infty left(x wedge yright) theta^2 e^{-theta(x+y)} , dy , dx\ &= int_{X_1(A)} int_0^x theta^2 ye^{-theta(x+y)} , dy , dx + int_{X_1(A)} int_x^infty theta^2 x e^{-theta(x+y)} , dy , dx \
&= int_{X_1(A)} left(-theta x e^{-2theta x} - e^{-2theta x} + e^{-theta x}right) , dx + int_{X_1(A)} theta x e^{-2theta x},dx \
&= int_{X_1(A)} left(e^{-theta x} - e^{-2theta x}right),dx = int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right)theta e^{-theta x},dx \
&= int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right) , dleft(mathbb P circ X_1^{-1}right)(x) = int_A frac 1 theta left(1-e^{-theta X_1}right) , dmathbb P \
&= mathbb Eleft[frac 1 theta left(1-e^{-theta X_1}right) mathbb 1_A right].
end{align*}
From this it follows that $boxed{mathbb Eleft[X_1 wedge X_2 | X_1right] = frac 1 theta left(1-e^{-theta X_1}right).}$
answered Jan 27 at 20:02
D FordD Ford
665313
$endgroup$
We're looking for a $sigma(X_1)$-measurable function $mathbb Eleft[X_1 wedge X_2 | X_1right] : Omega to mathbb R$ for which for every $A in sigma(X_1)$,
$$
mathbb Eleft[left(X_1 wedge X_2right) mathbb 1_Aright] = mathbb Eleft[mathbb Eleft[ X_1 wedge X_2 | X_1 right] mathbb 1_A right].
$$
Let $f_theta(x) = theta e^{-theta x}$ be the probability density of $X_1$ and of $X_2$, and let $lambda_n$ denote the $n$-dimensional Lebesgue measure. Then $f_theta , dlambda_1 = dleft(mathbb P circ X_1^{-1}right)$. Furthermore, by independence of $X_1$ and $X_2$, the joint probability density is $tilde f(x,y) = theta^2 e^{-theta(x+y)}$, and $dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) = tilde f , dlambda_2$, where $X_1 times X_2 : Omega to mathbb R^2$ is $omega mapsto left(X_1(omega), X_2(omega)right)$. Let $A in sigma(X_1)$. Then,
begin{align*}
mathbb Eleft[left(X_1 wedge X_2 right)mathbb 1_A right] &= int_A X_1 wedge X_2 , dmathbb P = int_{X_1(A) times [0,infty)} x wedge y , dleft(mathbb P circ left(X_1 times X_2 right)^{-1}right) \
&= int_{X_1(A)} int_0^infty left(x wedge yright) theta^2 e^{-theta(x+y)} , dy , dx\ &= int_{X_1(A)} int_0^x theta^2 ye^{-theta(x+y)} , dy , dx + int_{X_1(A)} int_x^infty theta^2 x e^{-theta(x+y)} , dy , dx \
&= int_{X_1(A)} left(-theta x e^{-2theta x} - e^{-2theta x} + e^{-theta x}right) , dx + int_{X_1(A)} theta x e^{-2theta x},dx \
&= int_{X_1(A)} left(e^{-theta x} - e^{-2theta x}right),dx = int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right)theta e^{-theta x},dx \
&= int_{X_1(A)} frac 1 theta left(1-e^{-theta x}right) , dleft(mathbb P circ X_1^{-1}right)(x) = int_A frac 1 theta left(1-e^{-theta X_1}right) , dmathbb P \
&= mathbb Eleft[frac 1 theta left(1-e^{-theta X_1}right) mathbb 1_A right].
end{align*}
From this it follows that $boxed{mathbb Eleft[X_1 wedge X_2 | X_1right] = frac 1 theta left(1-e^{-theta X_1}right).}$
answered Jan 27 at 20:02
D FordD Ford
665313
answered Jan 27 at 20:02
D FordD Ford
665313
answered Jan 27 at 20:02
D FordD Ford
665313
answered Jan 27 at 20:02
D FordD Ford
665313
665313
add a comment |
add a comment |
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$begingroup$
You cannot answer this without any independence assumption.
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– Kavi Rama Murthy
Jan 25 at 5:47
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Assuming independence, you might to try to find $mathbb E[k wedge X_2 ]$ for some constant $k ge 0$
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– Henry
Jan 25 at 8:49
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Ah yes, they are independent. I edited it.
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– D Ford
Jan 25 at 16:33
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@Henry I'm not sure how that helps if we assume independence. Is there an independence property you're invoking?
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– D Ford
Jan 25 at 16:34
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With independence, knowing $X_1=k$ tells you nothing new about the distribution of $X_2$, so my hint is helpful. Without independence, my suggestion would not help with the original question
$endgroup$
– Henry
Jan 25 at 16:41