Mixing
Given a joint density function, what is the conditional expectation $E[Y|x]$?
$begingroup$
The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
Evaluate the conditional expectation $E[Y|x]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.
My solution:
(My expected value is negative, which I'm pretty sure can never be the case. Also in line 4 and 8, are the domains correct?)
$$E[Y|x] = int_{-infty}^{infty}yf_{Y|X}(y|x)dy.$$
$$f_{Y|X}(y|x) = frac{f_{X,Y}(x,y)}{f_{X}(x)}.$$
$$f_{X}(x) = int_{-infty}^{infty}f_{X,Y}(x,y)dy$$
$$= int_{0}^{infty}e^{-y}dy $$
$$=left [ -e^{-y} right]Big|_0^infty $$
$$ = -e^{-infty}-(-e^{0})=1. $$
$$f_{X}(x) = 1, x geq 0.$$
Thus,
$$E[Y|x] = int_{0}^{infty}-ye^{-y}dy$$
$$= -int_{0}^{infty}ye^{-y}dy$$
After integration by parts,
$$E[Y|x] = left [ ye^{-y}+e^{-y} right]Big|_0^infty$$
$$=0-[e^{0} + 0]=-1$$
Therefore, $E[Y|x] = -1.$
Thanks in advance for reading this long problem and responding!
probability-theory probability-distributions conditional-expectation
asked Jan 23 at 7:40
beepboopbeepboopbeepboopbeepboop
10810
$endgroup$
add a comment |
$begingroup$
The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
Evaluate the conditional expectation $E[Y|x]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.
My solution:
(My expected value is negative, which I'm pretty sure can never be the case. Also in line 4 and 8, are the domains correct?)
$$E[Y|x] = int_{-infty}^{infty}yf_{Y|X}(y|x)dy.$$
$$f_{Y|X}(y|x) = frac{f_{X,Y}(x,y)}{f_{X}(x)}.$$
$$f_{X}(x) = int_{-infty}^{infty}f_{X,Y}(x,y)dy$$
$$= int_{0}^{infty}e^{-y}dy $$
$$=left [ -e^{-y} right]Big|_0^infty $$
$$ = -e^{-infty}-(-e^{0})=1. $$
$$f_{X}(x) = 1, x geq 0.$$
Thus,
$$E[Y|x] = int_{0}^{infty}-ye^{-y}dy$$
$$= -int_{0}^{infty}ye^{-y}dy$$
After integration by parts,
$$E[Y|x] = left [ ye^{-y}+e^{-y} right]Big|_0^infty$$
$$=0-[e^{0} + 0]=-1$$
Therefore, $E[Y|x] = -1.$
Thanks in advance for reading this long problem and responding!
probability-theory probability-distributions conditional-expectation
asked Jan 23 at 7:40
beepboopbeepboopbeepboopbeepboop
10810
$endgroup$
$begingroup$
As usual, your mistake is a consequence of the fact that you do not include the conditions on the joint PDF in the PDF itself. Here, the joint PDF $f_{X,Y}$, defined on the whole plane $mathbb R^2$, is $$f_{X,Y}(x,y)=e^{-y},mathbf 1_{0<x<y}$$ Using this correct formula in the following computations yields automatically the correct result. For example, starting from it, one gets $$f_X(x)=int_mathbb R f_{X,Y}(x,y)dy=mathbf 1_{x>0}int_x^infty e^{-y}dy=e^{-x}mathbf 1_{x>0}$$ Compare with your (flawed) computations.
$endgroup$
– Did
Jan 23 at 9:20
2
$begingroup$
Another point: the titles of your questions are systematically uninformative. Please stop the practice of choosing as title the first half of the first sentence of the post and choose instead a title describing what the question asks. For an example, see the revised title of the present question.
$endgroup$
– Did
Jan 23 at 9:30
$begingroup$
Why (re)introduce wrong notations in your title? FYI, $E(Ymid x)$ is meaningless.
$endgroup$
– Did
Jan 25 at 23:33
add a comment |
$begingroup$
The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
Evaluate the conditional expectation $E[Y|x]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.
My solution:
(My expected value is negative, which I'm pretty sure can never be the case. Also in line 4 and 8, are the domains correct?)
$$E[Y|x] = int_{-infty}^{infty}yf_{Y|X}(y|x)dy.$$
$$f_{Y|X}(y|x) = frac{f_{X,Y}(x,y)}{f_{X}(x)}.$$
$$f_{X}(x) = int_{-infty}^{infty}f_{X,Y}(x,y)dy$$
$$= int_{0}^{infty}e^{-y}dy $$
$$=left [ -e^{-y} right]Big|_0^infty $$
$$ = -e^{-infty}-(-e^{0})=1. $$
$$f_{X}(x) = 1, x geq 0.$$
Thus,
$$E[Y|x] = int_{0}^{infty}-ye^{-y}dy$$
$$= -int_{0}^{infty}ye^{-y}dy$$
After integration by parts,
$$E[Y|x] = left [ ye^{-y}+e^{-y} right]Big|_0^infty$$
$$=0-[e^{0} + 0]=-1$$
Therefore, $E[Y|x] = -1.$
Thanks in advance for reading this long problem and responding!
probability-theory probability-distributions conditional-expectation
asked Jan 23 at 7:40
beepboopbeepboopbeepboopbeepboop
10810
$endgroup$
The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
Evaluate the conditional expectation $E[Y|x]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.
My solution:
(My expected value is negative, which I'm pretty sure can never be the case. Also in line 4 and 8, are the domains correct?)
$$E[Y|x] = int_{-infty}^{infty}yf_{Y|X}(y|x)dy.$$
$$f_{Y|X}(y|x) = frac{f_{X,Y}(x,y)}{f_{X}(x)}.$$
$$f_{X}(x) = int_{-infty}^{infty}f_{X,Y}(x,y)dy$$
$$= int_{0}^{infty}e^{-y}dy $$
$$=left [ -e^{-y} right]Big|_0^infty $$
$$ = -e^{-infty}-(-e^{0})=1. $$
$$f_{X}(x) = 1, x geq 0.$$
Thus,
$$E[Y|x] = int_{0}^{infty}-ye^{-y}dy$$
$$= -int_{0}^{infty}ye^{-y}dy$$
After integration by parts,
$$E[Y|x] = left [ ye^{-y}+e^{-y} right]Big|_0^infty$$
$$=0-[e^{0} + 0]=-1$$
Therefore, $E[Y|x] = -1.$
Thanks in advance for reading this long problem and responding!
probability-theory probability-distributions conditional-expectation
probability-theory probability-distributions conditional-expectation
asked Jan 23 at 7:40
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 7:40
beepboopbeepboopbeepboopbeepboop
10810
edited Jan 25 at 20:26
beepboopbeepboop
asked Jan 23 at 7:40
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 7:40
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 7:40
beepboopbeepboopbeepboopbeepboop
10810
10810
$begingroup$
As usual, your mistake is a consequence of the fact that you do not include the conditions on the joint PDF in the PDF itself. Here, the joint PDF $f_{X,Y}$, defined on the whole plane $mathbb R^2$, is $$f_{X,Y}(x,y)=e^{-y},mathbf 1_{0<x<y}$$ Using this correct formula in the following computations yields automatically the correct result. For example, starting from it, one gets $$f_X(x)=int_mathbb R f_{X,Y}(x,y)dy=mathbf 1_{x>0}int_x^infty e^{-y}dy=e^{-x}mathbf 1_{x>0}$$ Compare with your (flawed) computations.
$endgroup$
– Did
Jan 23 at 9:20
2
$begingroup$
Another point: the titles of your questions are systematically uninformative. Please stop the practice of choosing as title the first half of the first sentence of the post and choose instead a title describing what the question asks. For an example, see the revised title of the present question.
$endgroup$
– Did
Jan 23 at 9:30
$begingroup$
Why (re)introduce wrong notations in your title? FYI, $E(Ymid x)$ is meaningless.
$endgroup$
– Did
Jan 25 at 23:33
add a comment |
$begingroup$
As usual, your mistake is a consequence of the fact that you do not include the conditions on the joint PDF in the PDF itself. Here, the joint PDF $f_{X,Y}$, defined on the whole plane $mathbb R^2$, is $$f_{X,Y}(x,y)=e^{-y},mathbf 1_{0<x<y}$$ Using this correct formula in the following computations yields automatically the correct result. For example, starting from it, one gets $$f_X(x)=int_mathbb R f_{X,Y}(x,y)dy=mathbf 1_{x>0}int_x^infty e^{-y}dy=e^{-x}mathbf 1_{x>0}$$ Compare with your (flawed) computations.
$endgroup$
– Did
Jan 23 at 9:20
2
$begingroup$
Another point: the titles of your questions are systematically uninformative. Please stop the practice of choosing as title the first half of the first sentence of the post and choose instead a title describing what the question asks. For an example, see the revised title of the present question.
$endgroup$
– Did
Jan 23 at 9:30
$begingroup$
Why (re)introduce wrong notations in your title? FYI, $E(Ymid x)$ is meaningless.
$endgroup$
– Did
Jan 25 at 23:33
$begingroup$
As usual, your mistake is a consequence of the fact that you do not include the conditions on the joint PDF in the PDF itself. Here, the joint PDF $f_{X,Y}$, defined on the whole plane $mathbb R^2$, is $$f_{X,Y}(x,y)=e^{-y},mathbf 1_{0<x<y}$$ Using this correct formula in the following computations yields automatically the correct result. For example, starting from it, one gets $$f_X(x)=int_mathbb R f_{X,Y}(x,y)dy=mathbf 1_{x>0}int_x^infty e^{-y}dy=e^{-x}mathbf 1_{x>0}$$ Compare with your (flawed) computations.
$endgroup$
– Did
Jan 23 at 9:20
$begingroup$
As usual, your mistake is a consequence of the fact that you do not include the conditions on the joint PDF in the PDF itself. Here, the joint PDF $f_{X,Y}$, defined on the whole plane $mathbb R^2$, is $$f_{X,Y}(x,y)=e^{-y},mathbf 1_{0<x<y}$$ Using this correct formula in the following computations yields automatically the correct result. For example, starting from it, one gets $$f_X(x)=int_mathbb R f_{X,Y}(x,y)dy=mathbf 1_{x>0}int_x^infty e^{-y}dy=e^{-x}mathbf 1_{x>0}$$ Compare with your (flawed) computations.
$endgroup$
– Did
Jan 23 at 9:20
2
2
$begingroup$
Another point: the titles of your questions are systematically uninformative. Please stop the practice of choosing as title the first half of the first sentence of the post and choose instead a title describing what the question asks. For an example, see the revised title of the present question.
$endgroup$
– Did
Jan 23 at 9:30
$begingroup$
Another point: the titles of your questions are systematically uninformative. Please stop the practice of choosing as title the first half of the first sentence of the post and choose instead a title describing what the question asks. For an example, see the revised title of the present question.
$endgroup$
– Did
Jan 23 at 9:30
$begingroup$
Why (re)introduce wrong notations in your title? FYI, $E(Ymid x)$ is meaningless.
$endgroup$
– Did
Jan 25 at 23:33
$begingroup$
Why (re)introduce wrong notations in your title? FYI, $E(Ymid x)$ is meaningless.
$endgroup$
– Did
Jan 25 at 23:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The limits in your fourth line are wrong since you know the joint density is $0$ unless $0leq x leq y le infty$ so it should be $$f_{X}(x) = int_{x}^{infty}e^{-y}dy , mbox{ } 0leq x le infty $$ leading to $$f_{X}(x)=
left{begin{matrix}e^{-x}, mbox{ } 0leq x le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
That will then give you
$$f_{Y mid X=x}(y)=
left{begin{matrix}dfrac{e^{-y}}{e^{-x}}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$ and $$E[Y mid X=x]=
int_{x}^{infty}yfrac{e^{-y}}{e^{-x}}dy = x+1$$
which is an example of the memorylessness property of the exponential distribution
answered Jan 23 at 8:03
HenryHenry
101k481168
$endgroup$
$begingroup$
Thank you, Henry!
$endgroup$
– beepboopbeepboop
Jan 23 at 19:54
add a comment |
$begingroup$
Yes, that negative expected value is absurd and a sign you did something wrong.
$$f_X(x) = int_0^{infty} e^{-y},dy$$
This line is the first mistake. The density function is zero unless $yge x$, so that should be $int_x^{infty}$ instead, and the probability you're dividing by will depend on $x$ - in a way that makes it a single-variable density. A constant density of $1$ on $[0,infty)$ for $f_X(x)$ is also an absurdity you could have caught.
$$E(Y|x) = int_0^infty -ye^{-y},dy$$
All right, this should be divided by what we just got for $f_X$ - but that's not the only problem. How, exactly, did a minus sign appear here out of nowhere? The integrand should be a probability density function times $y$, and densities are nonnegative.
Also, again, this will be an integral from $x$ to $infty$, as the density is zero unless $y ge x$.
Oh, and here's a neat interpretation of this distribution: Choose $x$ and $z$ independently from exponential distributions of parameter $1$, and then let $y=x+z$.
answered Jan 23 at 8:15
jmerryjmerry
13.6k1629
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
The limits in your fourth line are wrong since you know the joint density is $0$ unless $0leq x leq y le infty$ so it should be $$f_{X}(x) = int_{x}^{infty}e^{-y}dy , mbox{ } 0leq x le infty $$ leading to $$f_{X}(x)=
left{begin{matrix}e^{-x}, mbox{ } 0leq x le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
That will then give you
$$f_{Y mid X=x}(y)=
left{begin{matrix}dfrac{e^{-y}}{e^{-x}}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$ and $$E[Y mid X=x]=
int_{x}^{infty}yfrac{e^{-y}}{e^{-x}}dy = x+1$$
which is an example of the memorylessness property of the exponential distribution
answered Jan 23 at 8:03
HenryHenry
101k481168
$endgroup$
$begingroup$
Thank you, Henry!
$endgroup$
– beepboopbeepboop
Jan 23 at 19:54
add a comment |
$begingroup$
The limits in your fourth line are wrong since you know the joint density is $0$ unless $0leq x leq y le infty$ so it should be $$f_{X}(x) = int_{x}^{infty}e^{-y}dy , mbox{ } 0leq x le infty $$ leading to $$f_{X}(x)=
left{begin{matrix}e^{-x}, mbox{ } 0leq x le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
That will then give you
$$f_{Y mid X=x}(y)=
left{begin{matrix}dfrac{e^{-y}}{e^{-x}}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$ and $$E[Y mid X=x]=
int_{x}^{infty}yfrac{e^{-y}}{e^{-x}}dy = x+1$$
which is an example of the memorylessness property of the exponential distribution
answered Jan 23 at 8:03
HenryHenry
101k481168
$endgroup$
$begingroup$
Thank you, Henry!
$endgroup$
– beepboopbeepboop
Jan 23 at 19:54
add a comment |
$begingroup$
The limits in your fourth line are wrong since you know the joint density is $0$ unless $0leq x leq y le infty$ so it should be $$f_{X}(x) = int_{x}^{infty}e^{-y}dy , mbox{ } 0leq x le infty $$ leading to $$f_{X}(x)=
left{begin{matrix}e^{-x}, mbox{ } 0leq x le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
That will then give you
$$f_{Y mid X=x}(y)=
left{begin{matrix}dfrac{e^{-y}}{e^{-x}}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$ and $$E[Y mid X=x]=
int_{x}^{infty}yfrac{e^{-y}}{e^{-x}}dy = x+1$$
which is an example of the memorylessness property of the exponential distribution
answered Jan 23 at 8:03
HenryHenry
101k481168
$endgroup$
The limits in your fourth line are wrong since you know the joint density is $0$ unless $0leq x leq y le infty$ so it should be $$f_{X}(x) = int_{x}^{infty}e^{-y}dy , mbox{ } 0leq x le infty $$ leading to $$f_{X}(x)=
left{begin{matrix}e^{-x}, mbox{ } 0leq x le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
That will then give you
$$f_{Y mid X=x}(y)=
left{begin{matrix}dfrac{e^{-y}}{e^{-x}}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$ and $$E[Y mid X=x]=
int_{x}^{infty}yfrac{e^{-y}}{e^{-x}}dy = x+1$$
which is an example of the memorylessness property of the exponential distribution
answered Jan 23 at 8:03
HenryHenry
101k481168
edited Jan 23 at 8:15
answered Jan 23 at 8:03
HenryHenry
101k481168
answered Jan 23 at 8:03
HenryHenry
101k481168
answered Jan 23 at 8:03
HenryHenry
101k481168
101k481168
$begingroup$
Thank you, Henry!
$endgroup$
– beepboopbeepboop
Jan 23 at 19:54
add a comment |
$begingroup$
Thank you, Henry!
$endgroup$
– beepboopbeepboop
Jan 23 at 19:54
$begingroup$
Thank you, Henry!
$endgroup$
– beepboopbeepboop
Jan 23 at 19:54
$begingroup$
Thank you, Henry!
$endgroup$
– beepboopbeepboop
Jan 23 at 19:54
add a comment |
$begingroup$
Yes, that negative expected value is absurd and a sign you did something wrong.
$$f_X(x) = int_0^{infty} e^{-y},dy$$
This line is the first mistake. The density function is zero unless $yge x$, so that should be $int_x^{infty}$ instead, and the probability you're dividing by will depend on $x$ - in a way that makes it a single-variable density. A constant density of $1$ on $[0,infty)$ for $f_X(x)$ is also an absurdity you could have caught.
$$E(Y|x) = int_0^infty -ye^{-y},dy$$
All right, this should be divided by what we just got for $f_X$ - but that's not the only problem. How, exactly, did a minus sign appear here out of nowhere? The integrand should be a probability density function times $y$, and densities are nonnegative.
Also, again, this will be an integral from $x$ to $infty$, as the density is zero unless $y ge x$.
Oh, and here's a neat interpretation of this distribution: Choose $x$ and $z$ independently from exponential distributions of parameter $1$, and then let $y=x+z$.
answered Jan 23 at 8:15
jmerryjmerry
13.6k1629
$endgroup$
add a comment |
$begingroup$
Yes, that negative expected value is absurd and a sign you did something wrong.
$$f_X(x) = int_0^{infty} e^{-y},dy$$
This line is the first mistake. The density function is zero unless $yge x$, so that should be $int_x^{infty}$ instead, and the probability you're dividing by will depend on $x$ - in a way that makes it a single-variable density. A constant density of $1$ on $[0,infty)$ for $f_X(x)$ is also an absurdity you could have caught.
$$E(Y|x) = int_0^infty -ye^{-y},dy$$
All right, this should be divided by what we just got for $f_X$ - but that's not the only problem. How, exactly, did a minus sign appear here out of nowhere? The integrand should be a probability density function times $y$, and densities are nonnegative.
Also, again, this will be an integral from $x$ to $infty$, as the density is zero unless $y ge x$.
Oh, and here's a neat interpretation of this distribution: Choose $x$ and $z$ independently from exponential distributions of parameter $1$, and then let $y=x+z$.
answered Jan 23 at 8:15
jmerryjmerry
13.6k1629
$endgroup$
add a comment |
$begingroup$
Yes, that negative expected value is absurd and a sign you did something wrong.
$$f_X(x) = int_0^{infty} e^{-y},dy$$
This line is the first mistake. The density function is zero unless $yge x$, so that should be $int_x^{infty}$ instead, and the probability you're dividing by will depend on $x$ - in a way that makes it a single-variable density. A constant density of $1$ on $[0,infty)$ for $f_X(x)$ is also an absurdity you could have caught.
$$E(Y|x) = int_0^infty -ye^{-y},dy$$
All right, this should be divided by what we just got for $f_X$ - but that's not the only problem. How, exactly, did a minus sign appear here out of nowhere? The integrand should be a probability density function times $y$, and densities are nonnegative.
Also, again, this will be an integral from $x$ to $infty$, as the density is zero unless $y ge x$.
Oh, and here's a neat interpretation of this distribution: Choose $x$ and $z$ independently from exponential distributions of parameter $1$, and then let $y=x+z$.
answered Jan 23 at 8:15
jmerryjmerry
13.6k1629
$endgroup$
Yes, that negative expected value is absurd and a sign you did something wrong.
$$f_X(x) = int_0^{infty} e^{-y},dy$$
This line is the first mistake. The density function is zero unless $yge x$, so that should be $int_x^{infty}$ instead, and the probability you're dividing by will depend on $x$ - in a way that makes it a single-variable density. A constant density of $1$ on $[0,infty)$ for $f_X(x)$ is also an absurdity you could have caught.
$$E(Y|x) = int_0^infty -ye^{-y},dy$$
All right, this should be divided by what we just got for $f_X$ - but that's not the only problem. How, exactly, did a minus sign appear here out of nowhere? The integrand should be a probability density function times $y$, and densities are nonnegative.
Also, again, this will be an integral from $x$ to $infty$, as the density is zero unless $y ge x$.
Oh, and here's a neat interpretation of this distribution: Choose $x$ and $z$ independently from exponential distributions of parameter $1$, and then let $y=x+z$.
answered Jan 23 at 8:15
jmerryjmerry
13.6k1629
answered Jan 23 at 8:15
jmerryjmerry
13.6k1629
answered Jan 23 at 8:15
jmerryjmerry
13.6k1629
answered Jan 23 at 8:15
jmerryjmerry
13.6k1629
13.6k1629
add a comment |
add a comment |
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As usual, your mistake is a consequence of the fact that you do not include the conditions on the joint PDF in the PDF itself. Here, the joint PDF $f_{X,Y}$, defined on the whole plane $mathbb R^2$, is $$f_{X,Y}(x,y)=e^{-y},mathbf 1_{0<x<y}$$ Using this correct formula in the following computations yields automatically the correct result. For example, starting from it, one gets $$f_X(x)=int_mathbb R f_{X,Y}(x,y)dy=mathbf 1_{x>0}int_x^infty e^{-y}dy=e^{-x}mathbf 1_{x>0}$$ Compare with your (flawed) computations.
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– Did
Jan 23 at 9:20
2
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Another point: the titles of your questions are systematically uninformative. Please stop the practice of choosing as title the first half of the first sentence of the post and choose instead a title describing what the question asks. For an example, see the revised title of the present question.
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– Did
Jan 23 at 9:30
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Why (re)introduce wrong notations in your title? FYI, $E(Ymid x)$ is meaningless.
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– Did
Jan 25 at 23:33