Mixing
Expectation of a function of two independent random variables.
$begingroup$
Suppose X and Y are independent and uniformally distributed around [0, 1]. Define $Z = (X-Y)^2$. I'm interested in the variance $sigma_Z^2 = mathbb{E}[Z^2] - mathbb{E}[Z]^2$, and suppose I already know $mathbb{E}[Z]^2 = frac{1}{36}$.
For $mathbb{E}[Z^4]$, is computing $mathbb{E}[Z^2] cdot mathbb{E}[Z^2]$ sufficient enough since X and Y are i.i.d?
probability probability-theory statistics probability-distributions
asked Jan 21 at 21:40
TimelordViktoriousTimelordViktorious
487313
$endgroup$
add a comment |
$begingroup$
Suppose X and Y are independent and uniformally distributed around [0, 1]. Define $Z = (X-Y)^2$. I'm interested in the variance $sigma_Z^2 = mathbb{E}[Z^2] - mathbb{E}[Z]^2$, and suppose I already know $mathbb{E}[Z]^2 = frac{1}{36}$.
For $mathbb{E}[Z^4]$, is computing $mathbb{E}[Z^2] cdot mathbb{E}[Z^2]$ sufficient enough since X and Y are i.i.d?
probability probability-theory statistics probability-distributions
asked Jan 21 at 21:40
TimelordViktoriousTimelordViktorious
487313
$endgroup$
$begingroup$
Do you want the variance of $Z$ or $Z^2$? The formula you wrote is the variance of $Z^2$ but you have indexed $sigma^2$ by just $Z$. Which is it?
$endgroup$
– LoveTooNap29
Jan 21 at 21:52
$begingroup$
Sorry, there was a typo here.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
add a comment |
$begingroup$
Suppose X and Y are independent and uniformally distributed around [0, 1]. Define $Z = (X-Y)^2$. I'm interested in the variance $sigma_Z^2 = mathbb{E}[Z^2] - mathbb{E}[Z]^2$, and suppose I already know $mathbb{E}[Z]^2 = frac{1}{36}$.
For $mathbb{E}[Z^4]$, is computing $mathbb{E}[Z^2] cdot mathbb{E}[Z^2]$ sufficient enough since X and Y are i.i.d?
probability probability-theory statistics probability-distributions
asked Jan 21 at 21:40
TimelordViktoriousTimelordViktorious
487313
$endgroup$
Suppose X and Y are independent and uniformally distributed around [0, 1]. Define $Z = (X-Y)^2$. I'm interested in the variance $sigma_Z^2 = mathbb{E}[Z^2] - mathbb{E}[Z]^2$, and suppose I already know $mathbb{E}[Z]^2 = frac{1}{36}$.
For $mathbb{E}[Z^4]$, is computing $mathbb{E}[Z^2] cdot mathbb{E}[Z^2]$ sufficient enough since X and Y are i.i.d?
probability probability-theory statistics probability-distributions
probability probability-theory statistics probability-distributions
asked Jan 21 at 21:40
TimelordViktoriousTimelordViktorious
487313
asked Jan 21 at 21:40
TimelordViktoriousTimelordViktorious
487313
edited Jan 22 at 1:53
TimelordViktorious
asked Jan 21 at 21:40
TimelordViktoriousTimelordViktorious
487313
asked Jan 21 at 21:40
TimelordViktoriousTimelordViktorious
487313
asked Jan 21 at 21:40
TimelordViktoriousTimelordViktorious
487313
487313
$begingroup$
Do you want the variance of $Z$ or $Z^2$? The formula you wrote is the variance of $Z^2$ but you have indexed $sigma^2$ by just $Z$. Which is it?
$endgroup$
– LoveTooNap29
Jan 21 at 21:52
$begingroup$
Sorry, there was a typo here.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
add a comment |
$begingroup$
Do you want the variance of $Z$ or $Z^2$? The formula you wrote is the variance of $Z^2$ but you have indexed $sigma^2$ by just $Z$. Which is it?
$endgroup$
– LoveTooNap29
Jan 21 at 21:52
$begingroup$
Sorry, there was a typo here.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
Do you want the variance of $Z$ or $Z^2$? The formula you wrote is the variance of $Z^2$ but you have indexed $sigma^2$ by just $Z$. Which is it?
$endgroup$
– LoveTooNap29
Jan 21 at 21:52
$begingroup$
Do you want the variance of $Z$ or $Z^2$? The formula you wrote is the variance of $Z^2$ but you have indexed $sigma^2$ by just $Z$. Which is it?
$endgroup$
– LoveTooNap29
Jan 21 at 21:52
$begingroup$
Sorry, there was a typo here.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
Sorry, there was a typo here.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general the answer to "is $mathbb{E}(Z^4)=mathbb{E}(Z^2)^2$?" is no, so I would not try to use this. Just directly compute everything.
Assuming you want to compute the variance of $Z^2$:
Let $Z=(X-Y)^2$ where $X,Y$ are IID $mathcal{U}(0,1)$ RVs. The variance of $Z^2$ is given by
$$sigma_{Z^2}^2=mathbb{E}(Z^4)-mathbb{E}(Z^2)^2,$$
using the well known formula for variance. Both of these expectations can be computed directly. Lets start with the lower power one. Applying the binomial theorem,
$$Z^2=(X-Y)^4=X^4-4X^3Y+6X^2Y^2-4XY^3 +Y^4,$$
then taking expectations and using linearity and independence of $X$ and $Y$,
$$mathbb{E}(Z^2)=mathbb{E}(X^4)-4mathbb{E}(X^3)mathbb{E}(Y)+6mathbb{E}(X^2)mathbb{E}(Y^2)-4mathbb{E}(X)mathbb{E}(Y^3)+mathbb{E}(Y^4),$$
But since $X$ and $Y$ are also identically distributed uniform RVs on $(0,1)$, it follows that they have the same moments. So in turn this reduces to
$$mathbb{E}(Z^2)=2mathbb{E}(U^4)-8mathbb{E}(U^3)mathbb{E}(U)+6mathbb{E}(U^2)^2$$
where $U$ is any $mathcal{U}(0,1)$ RV. All of these can now be computed directly as standard integrals and once that is over with, just remember to square the result. Repeat the process for $Z^4$ and you are done after substituting both results back into the variance formula.
answered Jan 21 at 22:09
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
1
$begingroup$
FWIW, I don't think the expansion of Z^2 in the second step is entirely correct.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
@TimelordViktorious Ah! you are correct—I am off by some signs that unfortunately carry through the rest of the work. I'll edit accordingly.
$endgroup$
– LoveTooNap29
Jan 22 at 1:57
$begingroup$
@TimelordViktorious Should be good now but it's always nice to have a second set of eyes :)
$endgroup$
– LoveTooNap29
Jan 22 at 1:59
$begingroup$
Thanks a lot for your help!
$endgroup$
– TimelordViktorious
Jan 22 at 14:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In general the answer to "is $mathbb{E}(Z^4)=mathbb{E}(Z^2)^2$?" is no, so I would not try to use this. Just directly compute everything.
Assuming you want to compute the variance of $Z^2$:
Let $Z=(X-Y)^2$ where $X,Y$ are IID $mathcal{U}(0,1)$ RVs. The variance of $Z^2$ is given by
$$sigma_{Z^2}^2=mathbb{E}(Z^4)-mathbb{E}(Z^2)^2,$$
using the well known formula for variance. Both of these expectations can be computed directly. Lets start with the lower power one. Applying the binomial theorem,
$$Z^2=(X-Y)^4=X^4-4X^3Y+6X^2Y^2-4XY^3 +Y^4,$$
then taking expectations and using linearity and independence of $X$ and $Y$,
$$mathbb{E}(Z^2)=mathbb{E}(X^4)-4mathbb{E}(X^3)mathbb{E}(Y)+6mathbb{E}(X^2)mathbb{E}(Y^2)-4mathbb{E}(X)mathbb{E}(Y^3)+mathbb{E}(Y^4),$$
But since $X$ and $Y$ are also identically distributed uniform RVs on $(0,1)$, it follows that they have the same moments. So in turn this reduces to
$$mathbb{E}(Z^2)=2mathbb{E}(U^4)-8mathbb{E}(U^3)mathbb{E}(U)+6mathbb{E}(U^2)^2$$
where $U$ is any $mathcal{U}(0,1)$ RV. All of these can now be computed directly as standard integrals and once that is over with, just remember to square the result. Repeat the process for $Z^4$ and you are done after substituting both results back into the variance formula.
answered Jan 21 at 22:09
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
1
$begingroup$
FWIW, I don't think the expansion of Z^2 in the second step is entirely correct.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
@TimelordViktorious Ah! you are correct—I am off by some signs that unfortunately carry through the rest of the work. I'll edit accordingly.
$endgroup$
– LoveTooNap29
Jan 22 at 1:57
$begingroup$
@TimelordViktorious Should be good now but it's always nice to have a second set of eyes :)
$endgroup$
– LoveTooNap29
Jan 22 at 1:59
$begingroup$
Thanks a lot for your help!
$endgroup$
– TimelordViktorious
Jan 22 at 14:49
add a comment |
$begingroup$
In general the answer to "is $mathbb{E}(Z^4)=mathbb{E}(Z^2)^2$?" is no, so I would not try to use this. Just directly compute everything.
Assuming you want to compute the variance of $Z^2$:
Let $Z=(X-Y)^2$ where $X,Y$ are IID $mathcal{U}(0,1)$ RVs. The variance of $Z^2$ is given by
$$sigma_{Z^2}^2=mathbb{E}(Z^4)-mathbb{E}(Z^2)^2,$$
using the well known formula for variance. Both of these expectations can be computed directly. Lets start with the lower power one. Applying the binomial theorem,
$$Z^2=(X-Y)^4=X^4-4X^3Y+6X^2Y^2-4XY^3 +Y^4,$$
then taking expectations and using linearity and independence of $X$ and $Y$,
$$mathbb{E}(Z^2)=mathbb{E}(X^4)-4mathbb{E}(X^3)mathbb{E}(Y)+6mathbb{E}(X^2)mathbb{E}(Y^2)-4mathbb{E}(X)mathbb{E}(Y^3)+mathbb{E}(Y^4),$$
But since $X$ and $Y$ are also identically distributed uniform RVs on $(0,1)$, it follows that they have the same moments. So in turn this reduces to
$$mathbb{E}(Z^2)=2mathbb{E}(U^4)-8mathbb{E}(U^3)mathbb{E}(U)+6mathbb{E}(U^2)^2$$
where $U$ is any $mathcal{U}(0,1)$ RV. All of these can now be computed directly as standard integrals and once that is over with, just remember to square the result. Repeat the process for $Z^4$ and you are done after substituting both results back into the variance formula.
answered Jan 21 at 22:09
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
1
$begingroup$
FWIW, I don't think the expansion of Z^2 in the second step is entirely correct.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
@TimelordViktorious Ah! you are correct—I am off by some signs that unfortunately carry through the rest of the work. I'll edit accordingly.
$endgroup$
– LoveTooNap29
Jan 22 at 1:57
$begingroup$
@TimelordViktorious Should be good now but it's always nice to have a second set of eyes :)
$endgroup$
– LoveTooNap29
Jan 22 at 1:59
$begingroup$
Thanks a lot for your help!
$endgroup$
– TimelordViktorious
Jan 22 at 14:49
add a comment |
$begingroup$
In general the answer to "is $mathbb{E}(Z^4)=mathbb{E}(Z^2)^2$?" is no, so I would not try to use this. Just directly compute everything.
Assuming you want to compute the variance of $Z^2$:
Let $Z=(X-Y)^2$ where $X,Y$ are IID $mathcal{U}(0,1)$ RVs. The variance of $Z^2$ is given by
$$sigma_{Z^2}^2=mathbb{E}(Z^4)-mathbb{E}(Z^2)^2,$$
using the well known formula for variance. Both of these expectations can be computed directly. Lets start with the lower power one. Applying the binomial theorem,
$$Z^2=(X-Y)^4=X^4-4X^3Y+6X^2Y^2-4XY^3 +Y^4,$$
then taking expectations and using linearity and independence of $X$ and $Y$,
$$mathbb{E}(Z^2)=mathbb{E}(X^4)-4mathbb{E}(X^3)mathbb{E}(Y)+6mathbb{E}(X^2)mathbb{E}(Y^2)-4mathbb{E}(X)mathbb{E}(Y^3)+mathbb{E}(Y^4),$$
But since $X$ and $Y$ are also identically distributed uniform RVs on $(0,1)$, it follows that they have the same moments. So in turn this reduces to
$$mathbb{E}(Z^2)=2mathbb{E}(U^4)-8mathbb{E}(U^3)mathbb{E}(U)+6mathbb{E}(U^2)^2$$
where $U$ is any $mathcal{U}(0,1)$ RV. All of these can now be computed directly as standard integrals and once that is over with, just remember to square the result. Repeat the process for $Z^4$ and you are done after substituting both results back into the variance formula.
answered Jan 21 at 22:09
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
In general the answer to "is $mathbb{E}(Z^4)=mathbb{E}(Z^2)^2$?" is no, so I would not try to use this. Just directly compute everything.
Assuming you want to compute the variance of $Z^2$:
Let $Z=(X-Y)^2$ where $X,Y$ are IID $mathcal{U}(0,1)$ RVs. The variance of $Z^2$ is given by
$$sigma_{Z^2}^2=mathbb{E}(Z^4)-mathbb{E}(Z^2)^2,$$
using the well known formula for variance. Both of these expectations can be computed directly. Lets start with the lower power one. Applying the binomial theorem,
$$Z^2=(X-Y)^4=X^4-4X^3Y+6X^2Y^2-4XY^3 +Y^4,$$
then taking expectations and using linearity and independence of $X$ and $Y$,
$$mathbb{E}(Z^2)=mathbb{E}(X^4)-4mathbb{E}(X^3)mathbb{E}(Y)+6mathbb{E}(X^2)mathbb{E}(Y^2)-4mathbb{E}(X)mathbb{E}(Y^3)+mathbb{E}(Y^4),$$
But since $X$ and $Y$ are also identically distributed uniform RVs on $(0,1)$, it follows that they have the same moments. So in turn this reduces to
$$mathbb{E}(Z^2)=2mathbb{E}(U^4)-8mathbb{E}(U^3)mathbb{E}(U)+6mathbb{E}(U^2)^2$$
where $U$ is any $mathcal{U}(0,1)$ RV. All of these can now be computed directly as standard integrals and once that is over with, just remember to square the result. Repeat the process for $Z^4$ and you are done after substituting both results back into the variance formula.
answered Jan 21 at 22:09
LoveTooNap29LoveTooNap29
1,1331614
edited Jan 22 at 1:58
answered Jan 21 at 22:09
LoveTooNap29LoveTooNap29
1,1331614
answered Jan 21 at 22:09
LoveTooNap29LoveTooNap29
1,1331614
answered Jan 21 at 22:09
LoveTooNap29LoveTooNap29
1,1331614
1,1331614
1
$begingroup$
FWIW, I don't think the expansion of Z^2 in the second step is entirely correct.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
@TimelordViktorious Ah! you are correct—I am off by some signs that unfortunately carry through the rest of the work. I'll edit accordingly.
$endgroup$
– LoveTooNap29
Jan 22 at 1:57
$begingroup$
@TimelordViktorious Should be good now but it's always nice to have a second set of eyes :)
$endgroup$
– LoveTooNap29
Jan 22 at 1:59
$begingroup$
Thanks a lot for your help!
$endgroup$
– TimelordViktorious
Jan 22 at 14:49
add a comment |
1
$begingroup$
FWIW, I don't think the expansion of Z^2 in the second step is entirely correct.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
@TimelordViktorious Ah! you are correct—I am off by some signs that unfortunately carry through the rest of the work. I'll edit accordingly.
$endgroup$
– LoveTooNap29
Jan 22 at 1:57
$begingroup$
@TimelordViktorious Should be good now but it's always nice to have a second set of eyes :)
$endgroup$
– LoveTooNap29
Jan 22 at 1:59
$begingroup$
Thanks a lot for your help!
$endgroup$
– TimelordViktorious
Jan 22 at 14:49
1
1
$begingroup$
FWIW, I don't think the expansion of Z^2 in the second step is entirely correct.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
FWIW, I don't think the expansion of Z^2 in the second step is entirely correct.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52
$begingroup$
@TimelordViktorious Ah! you are correct—I am off by some signs that unfortunately carry through the rest of the work. I'll edit accordingly.
$endgroup$
– LoveTooNap29
Jan 22 at 1:57
$begingroup$
@TimelordViktorious Ah! you are correct—I am off by some signs that unfortunately carry through the rest of the work. I'll edit accordingly.
$endgroup$
– LoveTooNap29
Jan 22 at 1:57
$begingroup$
@TimelordViktorious Should be good now but it's always nice to have a second set of eyes :)
$endgroup$
– LoveTooNap29
Jan 22 at 1:59
$begingroup$
@TimelordViktorious Should be good now but it's always nice to have a second set of eyes :)
$endgroup$
– LoveTooNap29
Jan 22 at 1:59
$begingroup$
Thanks a lot for your help!
$endgroup$
– TimelordViktorious
Jan 22 at 14:49
$begingroup$
Thanks a lot for your help!
$endgroup$
– TimelordViktorious
Jan 22 at 14:49
add a comment |
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$begingroup$
Do you want the variance of $Z$ or $Z^2$? The formula you wrote is the variance of $Z^2$ but you have indexed $sigma^2$ by just $Z$. Which is it?
$endgroup$
– LoveTooNap29
Jan 21 at 21:52
$begingroup$
Sorry, there was a typo here.
$endgroup$
– TimelordViktorious
Jan 22 at 1:52