Mixing
Show that $e^{X^2/2} in L^1$ iff $e^{XY} in L^1$ iff $e^{|XY|} in L^1$
$begingroup$
let $X, Y$ be two identically distributed (both are $mathcal{N}(0,1)$) independent random variables
show that $e^{frac{X^2}{2}} in L^1 iff e^{XY} in L^1 iff e^{|XY|} in L^1$.
my attempt :
1st equivalence :
$$begin{align}
mathbb{E}[e^{XY}] &= frac{1}{2pi}int_{mathbb{R}}int_{mathbb{R}}e^{xy}e^{-frac{x^2}{2}}e^{-frac{y^2}{2}}dydx =frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}int_{mathbb{R}}e^{xy-frac{y^2}{2}}dydx \
&=frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}int_{mathbb{R}}e^{frac{x^2}{2}}e^{-frac{(x-y)^2}{2}}dydx \
&= frac{1}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}e^{-frac{x^2}{2}}int_{mathbb{R}}frac{1}{sqrt{2pi}}e^{-frac{u^2}{2}}dudx \
& = mathbb{E}[e^{frac{X^2}{2}}]
end{align} $$
I mean yeah this kinda proves that $e^{frac{X^2}{2}} in L^1 iff e^{XY} in L^1 $
but something is bothering me,
because $mathbb{E}[e^{frac{X^2}{2}}] = frac{1}{sqrt{2pi}}int_{mathbb{R}}dx = +infty$
Q1 :
can't we just say that $e^{frac{X^2}{2}} in L^1$ is a false claim therefore it can imply anything we desire ?
second equivalence : from the fact that $0< e^{XY} leq e^{|XY|}$
we conclude that $ e^{|XY|} in L^1 implies e^{XY} in L^1$
$$begin{align}
mathbb{E}[e^{|XY|}] &= frac{1}{2pi}int_{mathbb{R}}int_{mathbb{R}}e^{|xy|}e^{-frac{x^2}{2}}e^{-frac{y^2}{2}}dydx \
&=frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{|xy|}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{|xy|}e^{-frac{y^2}{2}}dy)dx \
&= frac{1}{2pi}[int_{0}^{+infty}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{xy}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{-xy}e^{-frac{y^2}{2}}dy)dx +int_{-infty}^{0}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{-xy}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{xy}e^{-frac{y^2}{2}}dy)dx]\
& leq text{Constant}[mathbb{E}[e^{XY}] + mathbb{E}[e^{-XY}] ] = text{Constant}_2[mathbb{E}[e^{XY}]]
end{align} $$
I used the fact that $-X$ and $Y$ are independant and that $X = -X, text{in distribution}$
Q2 :
was my attempt at proving 2nd equivalence correct ?
thanks !
edit 1 : pic of the original problem (it's in french) 
probability-theory lp-spaces expected-value
edited Feb 1 at 22:44
Did
249k23228466
asked Feb 1 at 17:44
rapidracimrapidracim
1,7441419
$endgroup$
|
show 10 more comments
$begingroup$
let $X, Y$ be two identically distributed (both are $mathcal{N}(0,1)$) independent random variables
show that $e^{frac{X^2}{2}} in L^1 iff e^{XY} in L^1 iff e^{|XY|} in L^1$.
my attempt :
1st equivalence :
$$begin{align}
mathbb{E}[e^{XY}] &= frac{1}{2pi}int_{mathbb{R}}int_{mathbb{R}}e^{xy}e^{-frac{x^2}{2}}e^{-frac{y^2}{2}}dydx =frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}int_{mathbb{R}}e^{xy-frac{y^2}{2}}dydx \
&=frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}int_{mathbb{R}}e^{frac{x^2}{2}}e^{-frac{(x-y)^2}{2}}dydx \
&= frac{1}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}e^{-frac{x^2}{2}}int_{mathbb{R}}frac{1}{sqrt{2pi}}e^{-frac{u^2}{2}}dudx \
& = mathbb{E}[e^{frac{X^2}{2}}]
end{align} $$
I mean yeah this kinda proves that $e^{frac{X^2}{2}} in L^1 iff e^{XY} in L^1 $
but something is bothering me,
because $mathbb{E}[e^{frac{X^2}{2}}] = frac{1}{sqrt{2pi}}int_{mathbb{R}}dx = +infty$
Q1 :
can't we just say that $e^{frac{X^2}{2}} in L^1$ is a false claim therefore it can imply anything we desire ?
second equivalence : from the fact that $0< e^{XY} leq e^{|XY|}$
we conclude that $ e^{|XY|} in L^1 implies e^{XY} in L^1$
$$begin{align}
mathbb{E}[e^{|XY|}] &= frac{1}{2pi}int_{mathbb{R}}int_{mathbb{R}}e^{|xy|}e^{-frac{x^2}{2}}e^{-frac{y^2}{2}}dydx \
&=frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{|xy|}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{|xy|}e^{-frac{y^2}{2}}dy)dx \
&= frac{1}{2pi}[int_{0}^{+infty}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{xy}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{-xy}e^{-frac{y^2}{2}}dy)dx +int_{-infty}^{0}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{-xy}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{xy}e^{-frac{y^2}{2}}dy)dx]\
& leq text{Constant}[mathbb{E}[e^{XY}] + mathbb{E}[e^{-XY}] ] = text{Constant}_2[mathbb{E}[e^{XY}]]
end{align} $$
I used the fact that $-X$ and $Y$ are independant and that $X = -X, text{in distribution}$
Q2 :
was my attempt at proving 2nd equivalence correct ?
thanks !
edit 1 : pic of the original problem (it's in french)
probability-theory lp-spaces expected-value
edited Feb 1 at 22:44
Did
249k23228466
asked Feb 1 at 17:44
rapidracimrapidracim
1,7441419
$endgroup$
1
$begingroup$
Are you sure you were asked to prove these (rather odd) equivalences, or did you make them up yourself, to solve another problem?
$endgroup$
– Did
Feb 1 at 21:29
$begingroup$
@Did here is a pic (imgur.com/a/5AgV1c8) of the problem, it's from an exam. calculer means 'compute', montrer que means 'show that', sans means 'without' and $M_x$ here denotes the moment generating function
$endgroup$
– rapidracim
Feb 1 at 21:36
$begingroup$
"Click below to consent to the use of this technology across the web" No thanks.
$endgroup$
– Did
Feb 1 at 22:18
$begingroup$
@Did link is messy ? I added the pic to my post for more convenience.
$endgroup$
– rapidracim
Feb 1 at 22:31
1
$begingroup$
Hmmm... Actually the whole exercise is wrong since, the random variable $e^{XY}$ failing to be integrable, the conditional expectation $E(e^{XY}mid X)$ does not exist.
$endgroup$
– Did
Feb 1 at 22:42
|
show 10 more comments
$begingroup$
let $X, Y$ be two identically distributed (both are $mathcal{N}(0,1)$) independent random variables
show that $e^{frac{X^2}{2}} in L^1 iff e^{XY} in L^1 iff e^{|XY|} in L^1$.
my attempt :
1st equivalence :
$$begin{align}
mathbb{E}[e^{XY}] &= frac{1}{2pi}int_{mathbb{R}}int_{mathbb{R}}e^{xy}e^{-frac{x^2}{2}}e^{-frac{y^2}{2}}dydx =frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}int_{mathbb{R}}e^{xy-frac{y^2}{2}}dydx \
&=frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}int_{mathbb{R}}e^{frac{x^2}{2}}e^{-frac{(x-y)^2}{2}}dydx \
&= frac{1}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}e^{-frac{x^2}{2}}int_{mathbb{R}}frac{1}{sqrt{2pi}}e^{-frac{u^2}{2}}dudx \
& = mathbb{E}[e^{frac{X^2}{2}}]
end{align} $$
I mean yeah this kinda proves that $e^{frac{X^2}{2}} in L^1 iff e^{XY} in L^1 $
but something is bothering me,
because $mathbb{E}[e^{frac{X^2}{2}}] = frac{1}{sqrt{2pi}}int_{mathbb{R}}dx = +infty$
Q1 :
can't we just say that $e^{frac{X^2}{2}} in L^1$ is a false claim therefore it can imply anything we desire ?
second equivalence : from the fact that $0< e^{XY} leq e^{|XY|}$
we conclude that $ e^{|XY|} in L^1 implies e^{XY} in L^1$
$$begin{align}
mathbb{E}[e^{|XY|}] &= frac{1}{2pi}int_{mathbb{R}}int_{mathbb{R}}e^{|xy|}e^{-frac{x^2}{2}}e^{-frac{y^2}{2}}dydx \
&=frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{|xy|}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{|xy|}e^{-frac{y^2}{2}}dy)dx \
&= frac{1}{2pi}[int_{0}^{+infty}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{xy}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{-xy}e^{-frac{y^2}{2}}dy)dx +int_{-infty}^{0}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{-xy}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{xy}e^{-frac{y^2}{2}}dy)dx]\
& leq text{Constant}[mathbb{E}[e^{XY}] + mathbb{E}[e^{-XY}] ] = text{Constant}_2[mathbb{E}[e^{XY}]]
end{align} $$
I used the fact that $-X$ and $Y$ are independant and that $X = -X, text{in distribution}$
Q2 :
was my attempt at proving 2nd equivalence correct ?
thanks !
edit 1 : pic of the original problem (it's in french)
probability-theory lp-spaces expected-value
edited Feb 1 at 22:44
Did
249k23228466
asked Feb 1 at 17:44
rapidracimrapidracim
1,7441419
$endgroup$
let $X, Y$ be two identically distributed (both are $mathcal{N}(0,1)$) independent random variables
show that $e^{frac{X^2}{2}} in L^1 iff e^{XY} in L^1 iff e^{|XY|} in L^1$.
my attempt :
1st equivalence :
$$begin{align}
mathbb{E}[e^{XY}] &= frac{1}{2pi}int_{mathbb{R}}int_{mathbb{R}}e^{xy}e^{-frac{x^2}{2}}e^{-frac{y^2}{2}}dydx =frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}int_{mathbb{R}}e^{xy-frac{y^2}{2}}dydx \
&=frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}int_{mathbb{R}}e^{frac{x^2}{2}}e^{-frac{(x-y)^2}{2}}dydx \
&= frac{1}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}e^{-frac{x^2}{2}}int_{mathbb{R}}frac{1}{sqrt{2pi}}e^{-frac{u^2}{2}}dudx \
& = mathbb{E}[e^{frac{X^2}{2}}]
end{align} $$
I mean yeah this kinda proves that $e^{frac{X^2}{2}} in L^1 iff e^{XY} in L^1 $
but something is bothering me,
because $mathbb{E}[e^{frac{X^2}{2}}] = frac{1}{sqrt{2pi}}int_{mathbb{R}}dx = +infty$
Q1 :
can't we just say that $e^{frac{X^2}{2}} in L^1$ is a false claim therefore it can imply anything we desire ?
second equivalence : from the fact that $0< e^{XY} leq e^{|XY|}$
we conclude that $ e^{|XY|} in L^1 implies e^{XY} in L^1$
$$begin{align}
mathbb{E}[e^{|XY|}] &= frac{1}{2pi}int_{mathbb{R}}int_{mathbb{R}}e^{|xy|}e^{-frac{x^2}{2}}e^{-frac{y^2}{2}}dydx \
&=frac{1}{2pi}int_{mathbb{R}}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{|xy|}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{|xy|}e^{-frac{y^2}{2}}dy)dx \
&= frac{1}{2pi}[int_{0}^{+infty}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{xy}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{-xy}e^{-frac{y^2}{2}}dy)dx +int_{-infty}^{0}e^{-frac{x^2}{2}}(int_{0}^{+infty}e^{-xy}e^{-frac{y^2}{2}}dy +int_{-infty}^{0}e^{xy}e^{-frac{y^2}{2}}dy)dx]\
& leq text{Constant}[mathbb{E}[e^{XY}] + mathbb{E}[e^{-XY}] ] = text{Constant}_2[mathbb{E}[e^{XY}]]
end{align} $$
I used the fact that $-X$ and $Y$ are independant and that $X = -X, text{in distribution}$
Q2 :
was my attempt at proving 2nd equivalence correct ?
thanks !
edit 1 : pic of the original problem (it's in french)
probability-theory lp-spaces expected-value
probability-theory lp-spaces expected-value
edited Feb 1 at 22:44
Did
249k23228466
asked Feb 1 at 17:44
rapidracimrapidracim
1,7441419
edited Feb 1 at 22:44
Did
249k23228466
asked Feb 1 at 17:44
rapidracimrapidracim
1,7441419
edited Feb 1 at 22:44
Did
249k23228466
edited Feb 1 at 22:44
Did
249k23228466
edited Feb 1 at 22:44
Did
249k23228466
249k23228466
asked Feb 1 at 17:44
rapidracimrapidracim
1,7441419
asked Feb 1 at 17:44
rapidracimrapidracim
1,7441419
asked Feb 1 at 17:44
rapidracimrapidracim
1,7441419
1,7441419
1
$begingroup$
Are you sure you were asked to prove these (rather odd) equivalences, or did you make them up yourself, to solve another problem?
$endgroup$
– Did
Feb 1 at 21:29
$begingroup$
@Did here is a pic (imgur.com/a/5AgV1c8) of the problem, it's from an exam. calculer means 'compute', montrer que means 'show that', sans means 'without' and $M_x$ here denotes the moment generating function
$endgroup$
– rapidracim
Feb 1 at 21:36
$begingroup$
"Click below to consent to the use of this technology across the web" No thanks.
$endgroup$
– Did
Feb 1 at 22:18
$begingroup$
@Did link is messy ? I added the pic to my post for more convenience.
$endgroup$
– rapidracim
Feb 1 at 22:31
1
$begingroup$
Hmmm... Actually the whole exercise is wrong since, the random variable $e^{XY}$ failing to be integrable, the conditional expectation $E(e^{XY}mid X)$ does not exist.
$endgroup$
– Did
Feb 1 at 22:42
|
show 10 more comments
1
$begingroup$
Are you sure you were asked to prove these (rather odd) equivalences, or did you make them up yourself, to solve another problem?
$endgroup$
– Did
Feb 1 at 21:29
$begingroup$
@Did here is a pic (imgur.com/a/5AgV1c8) of the problem, it's from an exam. calculer means 'compute', montrer que means 'show that', sans means 'without' and $M_x$ here denotes the moment generating function
$endgroup$
– rapidracim
Feb 1 at 21:36
$begingroup$
"Click below to consent to the use of this technology across the web" No thanks.
$endgroup$
– Did
Feb 1 at 22:18
$begingroup$
@Did link is messy ? I added the pic to my post for more convenience.
$endgroup$
– rapidracim
Feb 1 at 22:31
1
$begingroup$
Hmmm... Actually the whole exercise is wrong since, the random variable $e^{XY}$ failing to be integrable, the conditional expectation $E(e^{XY}mid X)$ does not exist.
$endgroup$
– Did
Feb 1 at 22:42
1
1
$begingroup$
Are you sure you were asked to prove these (rather odd) equivalences, or did you make them up yourself, to solve another problem?
$endgroup$
– Did
Feb 1 at 21:29
$begingroup$
Are you sure you were asked to prove these (rather odd) equivalences, or did you make them up yourself, to solve another problem?
$endgroup$
– Did
Feb 1 at 21:29
$begingroup$
@Did here is a pic (imgur.com/a/5AgV1c8) of the problem, it's from an exam. calculer means 'compute', montrer que means 'show that', sans means 'without' and $M_x$ here denotes the moment generating function
$endgroup$
– rapidracim
Feb 1 at 21:36
$begingroup$
@Did here is a pic (imgur.com/a/5AgV1c8) of the problem, it's from an exam. calculer means 'compute', montrer que means 'show that', sans means 'without' and $M_x$ here denotes the moment generating function
$endgroup$
– rapidracim
Feb 1 at 21:36
$begingroup$
"Click below to consent to the use of this technology across the web" No thanks.
$endgroup$
– Did
Feb 1 at 22:18
$begingroup$
"Click below to consent to the use of this technology across the web" No thanks.
$endgroup$
– Did
Feb 1 at 22:18
$begingroup$
@Did link is messy ? I added the pic to my post for more convenience.
$endgroup$
– rapidracim
Feb 1 at 22:31
$begingroup$
@Did link is messy ? I added the pic to my post for more convenience.
$endgroup$
– rapidracim
Feb 1 at 22:31
1
1
$begingroup$
Hmmm... Actually the whole exercise is wrong since, the random variable $e^{XY}$ failing to be integrable, the conditional expectation $E(e^{XY}mid X)$ does not exist.
$endgroup$
– Did
Feb 1 at 22:42
$begingroup$
Hmmm... Actually the whole exercise is wrong since, the random variable $e^{XY}$ failing to be integrable, the conditional expectation $E(e^{XY}mid X)$ does not exist.
$endgroup$
– Did
Feb 1 at 22:42
|
show 10 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$E[e^{XY}|X]$ exists since $0 leq e^{XY}$ (it's an extension from $L^2$ to non negative variables!!)
Let's suppose that we don't know the distribution of $X$. And suppose that $E[e^{frac{X^2}{2}}]<+infty$
Observe that:
$$frac{1}{sqrt{2pi}}int_{mathbb{R}}(int_{mathbb{R}}e^{|xy|}e^{-frac{1}{2}y^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]0;+infty[}e^{-frac{1}{2}(y-|x|)^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]-|x|;+infty[}e^{-frac{1}{2}u^2}du)dP_X(x) leq frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{mathbb{R}}e^{-frac{1}{2}u^2}du)dP_X(x) leq 2E[e^{frac{X^2}{2}}]<+infty$$
And then the equivalence!! (the exercise is true if the distribution of $X$ is unknown)
answered Feb 2 at 9:28
mathexmathex
1138
$endgroup$
$begingroup$
great detail concerning the distribution of $X$
$endgroup$
– rapidracim
Feb 2 at 9:35
$begingroup$
From where did you bring the exercise?
$endgroup$
– mathex
Feb 2 at 9:36
$begingroup$
We don't need to know the distribution of X. So the exercise is true!
$endgroup$
– mathex
Feb 2 at 9:38
$begingroup$
it's from an exam, the teacher said he found it in this book : amazon.fr/…
$endgroup$
– rapidracim
Feb 2 at 9:42
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$E[e^{XY}|X]$ exists since $0 leq e^{XY}$ (it's an extension from $L^2$ to non negative variables!!)
Let's suppose that we don't know the distribution of $X$. And suppose that $E[e^{frac{X^2}{2}}]<+infty$
Observe that:
$$frac{1}{sqrt{2pi}}int_{mathbb{R}}(int_{mathbb{R}}e^{|xy|}e^{-frac{1}{2}y^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]0;+infty[}e^{-frac{1}{2}(y-|x|)^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]-|x|;+infty[}e^{-frac{1}{2}u^2}du)dP_X(x) leq frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{mathbb{R}}e^{-frac{1}{2}u^2}du)dP_X(x) leq 2E[e^{frac{X^2}{2}}]<+infty$$
And then the equivalence!! (the exercise is true if the distribution of $X$ is unknown)
answered Feb 2 at 9:28
mathexmathex
1138
$endgroup$
$begingroup$
great detail concerning the distribution of $X$
$endgroup$
– rapidracim
Feb 2 at 9:35
$begingroup$
From where did you bring the exercise?
$endgroup$
– mathex
Feb 2 at 9:36
$begingroup$
We don't need to know the distribution of X. So the exercise is true!
$endgroup$
– mathex
Feb 2 at 9:38
$begingroup$
it's from an exam, the teacher said he found it in this book : amazon.fr/…
$endgroup$
– rapidracim
Feb 2 at 9:42
add a comment |
$begingroup$
$E[e^{XY}|X]$ exists since $0 leq e^{XY}$ (it's an extension from $L^2$ to non negative variables!!)
Let's suppose that we don't know the distribution of $X$. And suppose that $E[e^{frac{X^2}{2}}]<+infty$
Observe that:
$$frac{1}{sqrt{2pi}}int_{mathbb{R}}(int_{mathbb{R}}e^{|xy|}e^{-frac{1}{2}y^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]0;+infty[}e^{-frac{1}{2}(y-|x|)^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]-|x|;+infty[}e^{-frac{1}{2}u^2}du)dP_X(x) leq frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{mathbb{R}}e^{-frac{1}{2}u^2}du)dP_X(x) leq 2E[e^{frac{X^2}{2}}]<+infty$$
And then the equivalence!! (the exercise is true if the distribution of $X$ is unknown)
answered Feb 2 at 9:28
mathexmathex
1138
$endgroup$
$begingroup$
great detail concerning the distribution of $X$
$endgroup$
– rapidracim
Feb 2 at 9:35
$begingroup$
From where did you bring the exercise?
$endgroup$
– mathex
Feb 2 at 9:36
$begingroup$
We don't need to know the distribution of X. So the exercise is true!
$endgroup$
– mathex
Feb 2 at 9:38
$begingroup$
it's from an exam, the teacher said he found it in this book : amazon.fr/…
$endgroup$
– rapidracim
Feb 2 at 9:42
add a comment |
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$E[e^{XY}|X]$ exists since $0 leq e^{XY}$ (it's an extension from $L^2$ to non negative variables!!)
Let's suppose that we don't know the distribution of $X$. And suppose that $E[e^{frac{X^2}{2}}]<+infty$
Observe that:
$$frac{1}{sqrt{2pi}}int_{mathbb{R}}(int_{mathbb{R}}e^{|xy|}e^{-frac{1}{2}y^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]0;+infty[}e^{-frac{1}{2}(y-|x|)^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]-|x|;+infty[}e^{-frac{1}{2}u^2}du)dP_X(x) leq frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{mathbb{R}}e^{-frac{1}{2}u^2}du)dP_X(x) leq 2E[e^{frac{X^2}{2}}]<+infty$$
And then the equivalence!! (the exercise is true if the distribution of $X$ is unknown)
answered Feb 2 at 9:28
mathexmathex
1138
$endgroup$
$E[e^{XY}|X]$ exists since $0 leq e^{XY}$ (it's an extension from $L^2$ to non negative variables!!)
Let's suppose that we don't know the distribution of $X$. And suppose that $E[e^{frac{X^2}{2}}]<+infty$
Observe that:
$$frac{1}{sqrt{2pi}}int_{mathbb{R}}(int_{mathbb{R}}e^{|xy|}e^{-frac{1}{2}y^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]0;+infty[}e^{-frac{1}{2}(y-|x|)^2}dy)dP_X(x)=frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{]-|x|;+infty[}e^{-frac{1}{2}u^2}du)dP_X(x) leq frac{2}{sqrt{2pi}}int_{mathbb{R}}e^{frac{x^2}{2}}(int_{mathbb{R}}e^{-frac{1}{2}u^2}du)dP_X(x) leq 2E[e^{frac{X^2}{2}}]<+infty$$
And then the equivalence!! (the exercise is true if the distribution of $X$ is unknown)
answered Feb 2 at 9:28
mathexmathex
1138
answered Feb 2 at 9:28
mathexmathex
1138
answered Feb 2 at 9:28
mathexmathex
1138
answered Feb 2 at 9:28
mathexmathex
1138
1138
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great detail concerning the distribution of $X$
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– rapidracim
Feb 2 at 9:35
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From where did you bring the exercise?
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– mathex
Feb 2 at 9:36
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We don't need to know the distribution of X. So the exercise is true!
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– mathex
Feb 2 at 9:38
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it's from an exam, the teacher said he found it in this book : amazon.fr/…
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– rapidracim
Feb 2 at 9:42
add a comment |
$begingroup$
great detail concerning the distribution of $X$
$endgroup$
– rapidracim
Feb 2 at 9:35
$begingroup$
From where did you bring the exercise?
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– mathex
Feb 2 at 9:36
$begingroup$
We don't need to know the distribution of X. So the exercise is true!
$endgroup$
– mathex
Feb 2 at 9:38
$begingroup$
it's from an exam, the teacher said he found it in this book : amazon.fr/…
$endgroup$
– rapidracim
Feb 2 at 9:42
$begingroup$
great detail concerning the distribution of $X$
$endgroup$
– rapidracim
Feb 2 at 9:35
$begingroup$
great detail concerning the distribution of $X$
$endgroup$
– rapidracim
Feb 2 at 9:35
$begingroup$
From where did you bring the exercise?
$endgroup$
– mathex
Feb 2 at 9:36
$begingroup$
From where did you bring the exercise?
$endgroup$
– mathex
Feb 2 at 9:36
$begingroup$
We don't need to know the distribution of X. So the exercise is true!
$endgroup$
– mathex
Feb 2 at 9:38
$begingroup$
We don't need to know the distribution of X. So the exercise is true!
$endgroup$
– mathex
Feb 2 at 9:38
$begingroup$
it's from an exam, the teacher said he found it in this book : amazon.fr/…
$endgroup$
– rapidracim
Feb 2 at 9:42
$begingroup$
it's from an exam, the teacher said he found it in this book : amazon.fr/…
$endgroup$
– rapidracim
Feb 2 at 9:42
add a comment |
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1
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Are you sure you were asked to prove these (rather odd) equivalences, or did you make them up yourself, to solve another problem?
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– Did
Feb 1 at 21:29
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@Did here is a pic (imgur.com/a/5AgV1c8) of the problem, it's from an exam. calculer means 'compute', montrer que means 'show that', sans means 'without' and $M_x$ here denotes the moment generating function
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– rapidracim
Feb 1 at 21:36
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"Click below to consent to the use of this technology across the web" No thanks.
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– Did
Feb 1 at 22:18
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@Did link is messy ? I added the pic to my post for more convenience.
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– rapidracim
Feb 1 at 22:31
1
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Hmmm... Actually the whole exercise is wrong since, the random variable $e^{XY}$ failing to be integrable, the conditional expectation $E(e^{XY}mid X)$ does not exist.
$endgroup$
– Did
Feb 1 at 22:42