Distribution of the difference of two normal random variables. The 2019 Stack Overflow...
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Distribution of the difference of two normal random variables.
The 2019 Stack Overflow Developer Survey Results Are InDistribution function of X-Y for normally distributed random variablesFinding the pdf of the squared difference between two independent standard normal random variablesRandom variables $X,Y$ such that $E(X|Y)=E(Y|X)$ a.s.Probability involving dependent normal variablesJoint distribution of the sum and product of two i.i.d. centered normal random variablesMoment generating function of a random number of IIDs?Independent, Identically Distributed Sequence of Poisson Random Variables: Approximating ProbabilityFind the median of a function of a normal random variable.Deriving the distribution of poisson random variables.Showing convergence of a random variable in distribution to a standard normal random variableWhy convert to a T distribution rather than use the standard normal distribution?Distribution of Product of Arbitrary Number of Random VariablesFinding the Probability from the sum of 3 random variables
$begingroup$
If $U$ and $V$ are independent identically distributed standard normal, what is the distribution of their difference?
I will present my answer here. I am hoping to know if I am right or wrong.
Using the method of moment generating functions, we have
begin{align*}
M_{U-V}(t)&=Eleft[e^{t(U-V)}right]\
&=Eleft[e^{tU}right]Eleft[e^{tV}right]\
&=M_U(t)M_V(t)\
&=left(M_U(t)right)^2\
&=left(e^{mu t+frac{1}{2}t^2sigma ^2}right)^2\
&=e^{2mu t+t^2sigma ^2}\
end{align*}
The last expression is the moment generating function for a random variable distributed normal with mean $2mu$ and variance $2sigma ^2$. Thus $U-Vsim N(2mu,2sigma ^2)$.
For the third line from the bottom, it follows from the fact that the moment generating functions are identical for $U$ and $V$.
Thanks for your input.
EDIT: OH I already see that I made a mistake, since the random variables are distributed STANDARD normal. I will change my answer to say $U-Vsim N(0,2)$.
probability statistics moment-generating-functions
asked Sep 2 '14 at 17:46
nonremovablenonremovable
7781619
$endgroup$
|
show 1 more comment
$begingroup$
If $U$ and $V$ are independent identically distributed standard normal, what is the distribution of their difference?
I will present my answer here. I am hoping to know if I am right or wrong.
Using the method of moment generating functions, we have
begin{align*}
M_{U-V}(t)&=Eleft[e^{t(U-V)}right]\
&=Eleft[e^{tU}right]Eleft[e^{tV}right]\
&=M_U(t)M_V(t)\
&=left(M_U(t)right)^2\
&=left(e^{mu t+frac{1}{2}t^2sigma ^2}right)^2\
&=e^{2mu t+t^2sigma ^2}\
end{align*}
The last expression is the moment generating function for a random variable distributed normal with mean $2mu$ and variance $2sigma ^2$. Thus $U-Vsim N(2mu,2sigma ^2)$.
For the third line from the bottom, it follows from the fact that the moment generating functions are identical for $U$ and $V$.
Thanks for your input.
EDIT: OH I already see that I made a mistake, since the random variables are distributed STANDARD normal. I will change my answer to say $U-Vsim N(0,2)$.
probability statistics moment-generating-functions
asked Sep 2 '14 at 17:46
nonremovablenonremovable
7781619
$endgroup$
2
$begingroup$
I think you made a sign error somewhere. The mean of $U-V$ should be zero even if $U$ and $V$ have nonzero mean $mu$. Shouldn't your second line be $E[e^{tU}]E[e^{-tV}]$?
$endgroup$
– Bungo
Sep 2 '14 at 17:54
$begingroup$
Ah, yes it should. Thank you.
$endgroup$
– nonremovable
Sep 2 '14 at 17:55
$begingroup$
Aside from that, your solution looks fine.
$endgroup$
– Bungo
Sep 2 '14 at 17:56
$begingroup$
Yeah, I changed the wrong sign, but in the end the answer still came out to $N(0,2)$.
$endgroup$
– nonremovable
Sep 2 '14 at 17:58
5
$begingroup$
Having $$E[U - V] = E[U] - E[V] = mu_U - mu_V$$ and $$Var(U - V) = Var(U) + Var(V) = sigma_U^2 + sigma_V^2$$ then $$(U - V) sim N(mu_U - mu_V, sigma_U^2 + sigma_V^2)$$
$endgroup$
– Dor
Sep 13 '15 at 8:57
|
show 1 more comment
$begingroup$
If $U$ and $V$ are independent identically distributed standard normal, what is the distribution of their difference?
I will present my answer here. I am hoping to know if I am right or wrong.
Using the method of moment generating functions, we have
begin{align*}
M_{U-V}(t)&=Eleft[e^{t(U-V)}right]\
&=Eleft[e^{tU}right]Eleft[e^{tV}right]\
&=M_U(t)M_V(t)\
&=left(M_U(t)right)^2\
&=left(e^{mu t+frac{1}{2}t^2sigma ^2}right)^2\
&=e^{2mu t+t^2sigma ^2}\
end{align*}
The last expression is the moment generating function for a random variable distributed normal with mean $2mu$ and variance $2sigma ^2$. Thus $U-Vsim N(2mu,2sigma ^2)$.
For the third line from the bottom, it follows from the fact that the moment generating functions are identical for $U$ and $V$.
Thanks for your input.
EDIT: OH I already see that I made a mistake, since the random variables are distributed STANDARD normal. I will change my answer to say $U-Vsim N(0,2)$.
probability statistics moment-generating-functions
asked Sep 2 '14 at 17:46
nonremovablenonremovable
7781619
$endgroup$
If $U$ and $V$ are independent identically distributed standard normal, what is the distribution of their difference?
I will present my answer here. I am hoping to know if I am right or wrong.
Using the method of moment generating functions, we have
begin{align*}
M_{U-V}(t)&=Eleft[e^{t(U-V)}right]\
&=Eleft[e^{tU}right]Eleft[e^{tV}right]\
&=M_U(t)M_V(t)\
&=left(M_U(t)right)^2\
&=left(e^{mu t+frac{1}{2}t^2sigma ^2}right)^2\
&=e^{2mu t+t^2sigma ^2}\
end{align*}
The last expression is the moment generating function for a random variable distributed normal with mean $2mu$ and variance $2sigma ^2$. Thus $U-Vsim N(2mu,2sigma ^2)$.
For the third line from the bottom, it follows from the fact that the moment generating functions are identical for $U$ and $V$.
Thanks for your input.
EDIT: OH I already see that I made a mistake, since the random variables are distributed STANDARD normal. I will change my answer to say $U-Vsim N(0,2)$.
probability statistics moment-generating-functions
probability statistics moment-generating-functions
asked Sep 2 '14 at 17:46
nonremovablenonremovable
7781619
asked Sep 2 '14 at 17:46
nonremovablenonremovable
7781619
asked Sep 2 '14 at 17:46
nonremovablenonremovable
7781619
asked Sep 2 '14 at 17:46
nonremovablenonremovable
7781619
asked Sep 2 '14 at 17:46
nonremovablenonremovable
7781619
7781619
2
$begingroup$
I think you made a sign error somewhere. The mean of $U-V$ should be zero even if $U$ and $V$ have nonzero mean $mu$. Shouldn't your second line be $E[e^{tU}]E[e^{-tV}]$?
$endgroup$
– Bungo
Sep 2 '14 at 17:54
$begingroup$
Ah, yes it should. Thank you.
$endgroup$
– nonremovable
Sep 2 '14 at 17:55
$begingroup$
Aside from that, your solution looks fine.
$endgroup$
– Bungo
Sep 2 '14 at 17:56
$begingroup$
Yeah, I changed the wrong sign, but in the end the answer still came out to $N(0,2)$.
$endgroup$
– nonremovable
Sep 2 '14 at 17:58
5
$begingroup$
Having $$E[U - V] = E[U] - E[V] = mu_U - mu_V$$ and $$Var(U - V) = Var(U) + Var(V) = sigma_U^2 + sigma_V^2$$ then $$(U - V) sim N(mu_U - mu_V, sigma_U^2 + sigma_V^2)$$
$endgroup$
– Dor
Sep 13 '15 at 8:57
|
show 1 more comment
2
$begingroup$
I think you made a sign error somewhere. The mean of $U-V$ should be zero even if $U$ and $V$ have nonzero mean $mu$. Shouldn't your second line be $E[e^{tU}]E[e^{-tV}]$?
$endgroup$
– Bungo
Sep 2 '14 at 17:54
$begingroup$
Ah, yes it should. Thank you.
$endgroup$
– nonremovable
Sep 2 '14 at 17:55
$begingroup$
Aside from that, your solution looks fine.
$endgroup$
– Bungo
Sep 2 '14 at 17:56
$begingroup$
Yeah, I changed the wrong sign, but in the end the answer still came out to $N(0,2)$.
$endgroup$
– nonremovable
Sep 2 '14 at 17:58
5
$begingroup$
Having $$E[U - V] = E[U] - E[V] = mu_U - mu_V$$ and $$Var(U - V) = Var(U) + Var(V) = sigma_U^2 + sigma_V^2$$ then $$(U - V) sim N(mu_U - mu_V, sigma_U^2 + sigma_V^2)$$
$endgroup$
– Dor
Sep 13 '15 at 8:57
2
2
$begingroup$
I think you made a sign error somewhere. The mean of $U-V$ should be zero even if $U$ and $V$ have nonzero mean $mu$. Shouldn't your second line be $E[e^{tU}]E[e^{-tV}]$?
$endgroup$
– Bungo
Sep 2 '14 at 17:54
$begingroup$
I think you made a sign error somewhere. The mean of $U-V$ should be zero even if $U$ and $V$ have nonzero mean $mu$. Shouldn't your second line be $E[e^{tU}]E[e^{-tV}]$?
$endgroup$
– Bungo
Sep 2 '14 at 17:54
$begingroup$
Ah, yes it should. Thank you.
$endgroup$
– nonremovable
Sep 2 '14 at 17:55
$begingroup$
Ah, yes it should. Thank you.
$endgroup$
– nonremovable
Sep 2 '14 at 17:55
$begingroup$
Aside from that, your solution looks fine.
$endgroup$
– Bungo
Sep 2 '14 at 17:56
$begingroup$
Aside from that, your solution looks fine.
$endgroup$
– Bungo
Sep 2 '14 at 17:56
$begingroup$
Yeah, I changed the wrong sign, but in the end the answer still came out to $N(0,2)$.
$endgroup$
– nonremovable
Sep 2 '14 at 17:58
$begingroup$
Yeah, I changed the wrong sign, but in the end the answer still came out to $N(0,2)$.
$endgroup$
– nonremovable
Sep 2 '14 at 17:58
5
5
$begingroup$
Having $$E[U - V] = E[U] - E[V] = mu_U - mu_V$$ and $$Var(U - V) = Var(U) + Var(V) = sigma_U^2 + sigma_V^2$$ then $$(U - V) sim N(mu_U - mu_V, sigma_U^2 + sigma_V^2)$$
$endgroup$
– Dor
Sep 13 '15 at 8:57
$begingroup$
Having $$E[U - V] = E[U] - E[V] = mu_U - mu_V$$ and $$Var(U - V) = Var(U) + Var(V) = sigma_U^2 + sigma_V^2$$ then $$(U - V) sim N(mu_U - mu_V, sigma_U^2 + sigma_V^2)$$
$endgroup$
– Dor
Sep 13 '15 at 8:57
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
In addition to the solution by the OP using the moment generating function, I'll provide a (nearly trivial) solution when the rules about the sum and linear transformations of normal distributions are known.
The distribution of $U-V$ is identical to $U+a cdot V$ with $a=-1$. So from the cited rules we know that $U+Vcdot a sim N(mu_U + acdot mu_V,~sigma_U^2 + a^2 cdot sigma_V^2) = N(mu_U - mu_V,~sigma_U^2 + sigma_V^2)~ text{(for $a = -1$)} = N(0,~2)~text{(for standard normal distributed variables)}$.
Edit 2017-11-20: After I rejected the correction proposed by @Sheljohn of the variance and one typo, several times, he wrote them in a comment, so I finally did see them. Thank you @Sheljohn!
answered Dec 8 '16 at 11:37
QaswedQaswed
311313
$endgroup$
$begingroup$
If $U$ and $V$ were not independent, would $sigma_{U+V}^2$ be equal to $sigma_U^2+sigma_V^2+2rhosigma_Usigma_V$ where $rho$ is correlation?
$endgroup$
– Milos
Jun 18 '17 at 21:59
2
$begingroup$
@Qaswed -1: $U+aV$ is not distributed as $mathcal{N}( mu_U + amu V, sigma_U^2 + |a| sigma_V^2 )$; $mu_U + amu V$ makes no sense, and the variance is $sigma_U^2 + a^2 sigma_V^2$
$endgroup$
– Sheljohn
Nov 18 '17 at 20:41
2
$begingroup$
@Sheljohn you are right: $a cdot mu V$ is a typo and should be $a cdot mu_V$. And for the variance part it should be $a^2$ instead of $|a|$. I reject the edits as I only thought they are only changes of style. Sorry, my bad!
$endgroup$
– Qaswed
Nov 20 '17 at 7:24
add a comment |
$begingroup$
The currently upvoted answer is wrong, and the author rejected attempts to edit despite 6 reviewers' approval. So here it is; if one knows the rules about the sum and linear transformations of normal distributions, then the distribution of $U-V$ is:
$$
U-V sim U + aV sim mathcal{N}big( mu_U + amu_V, sigma_U^2 + a^2sigma_V^2 big) = mathcal{N}big( mu_U - mu_V, sigma_U^2 + sigma_V^2 big)
$$
where $a=-1$ and $(mu,sigma)$ denote the mean and std for each variable.
answered Nov 18 '17 at 20:44
SheljohnSheljohn
1,291823
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In addition to the solution by the OP using the moment generating function, I'll provide a (nearly trivial) solution when the rules about the sum and linear transformations of normal distributions are known.
The distribution of $U-V$ is identical to $U+a cdot V$ with $a=-1$. So from the cited rules we know that $U+Vcdot a sim N(mu_U + acdot mu_V,~sigma_U^2 + a^2 cdot sigma_V^2) = N(mu_U - mu_V,~sigma_U^2 + sigma_V^2)~ text{(for $a = -1$)} = N(0,~2)~text{(for standard normal distributed variables)}$.
Edit 2017-11-20: After I rejected the correction proposed by @Sheljohn of the variance and one typo, several times, he wrote them in a comment, so I finally did see them. Thank you @Sheljohn!
answered Dec 8 '16 at 11:37
QaswedQaswed
311313
$endgroup$
$begingroup$
If $U$ and $V$ were not independent, would $sigma_{U+V}^2$ be equal to $sigma_U^2+sigma_V^2+2rhosigma_Usigma_V$ where $rho$ is correlation?
$endgroup$
– Milos
Jun 18 '17 at 21:59
2
$begingroup$
@Qaswed -1: $U+aV$ is not distributed as $mathcal{N}( mu_U + amu V, sigma_U^2 + |a| sigma_V^2 )$; $mu_U + amu V$ makes no sense, and the variance is $sigma_U^2 + a^2 sigma_V^2$
$endgroup$
– Sheljohn
Nov 18 '17 at 20:41
2
$begingroup$
@Sheljohn you are right: $a cdot mu V$ is a typo and should be $a cdot mu_V$. And for the variance part it should be $a^2$ instead of $|a|$. I reject the edits as I only thought they are only changes of style. Sorry, my bad!
$endgroup$
– Qaswed
Nov 20 '17 at 7:24
add a comment |
$begingroup$
In addition to the solution by the OP using the moment generating function, I'll provide a (nearly trivial) solution when the rules about the sum and linear transformations of normal distributions are known.
The distribution of $U-V$ is identical to $U+a cdot V$ with $a=-1$. So from the cited rules we know that $U+Vcdot a sim N(mu_U + acdot mu_V,~sigma_U^2 + a^2 cdot sigma_V^2) = N(mu_U - mu_V,~sigma_U^2 + sigma_V^2)~ text{(for $a = -1$)} = N(0,~2)~text{(for standard normal distributed variables)}$.
Edit 2017-11-20: After I rejected the correction proposed by @Sheljohn of the variance and one typo, several times, he wrote them in a comment, so I finally did see them. Thank you @Sheljohn!
answered Dec 8 '16 at 11:37
QaswedQaswed
311313
$endgroup$
$begingroup$
If $U$ and $V$ were not independent, would $sigma_{U+V}^2$ be equal to $sigma_U^2+sigma_V^2+2rhosigma_Usigma_V$ where $rho$ is correlation?
$endgroup$
– Milos
Jun 18 '17 at 21:59
2
$begingroup$
@Qaswed -1: $U+aV$ is not distributed as $mathcal{N}( mu_U + amu V, sigma_U^2 + |a| sigma_V^2 )$; $mu_U + amu V$ makes no sense, and the variance is $sigma_U^2 + a^2 sigma_V^2$
$endgroup$
– Sheljohn
Nov 18 '17 at 20:41
2
$begingroup$
@Sheljohn you are right: $a cdot mu V$ is a typo and should be $a cdot mu_V$. And for the variance part it should be $a^2$ instead of $|a|$. I reject the edits as I only thought they are only changes of style. Sorry, my bad!
$endgroup$
– Qaswed
Nov 20 '17 at 7:24
add a comment |
$begingroup$
In addition to the solution by the OP using the moment generating function, I'll provide a (nearly trivial) solution when the rules about the sum and linear transformations of normal distributions are known.
The distribution of $U-V$ is identical to $U+a cdot V$ with $a=-1$. So from the cited rules we know that $U+Vcdot a sim N(mu_U + acdot mu_V,~sigma_U^2 + a^2 cdot sigma_V^2) = N(mu_U - mu_V,~sigma_U^2 + sigma_V^2)~ text{(for $a = -1$)} = N(0,~2)~text{(for standard normal distributed variables)}$.
Edit 2017-11-20: After I rejected the correction proposed by @Sheljohn of the variance and one typo, several times, he wrote them in a comment, so I finally did see them. Thank you @Sheljohn!
answered Dec 8 '16 at 11:37
QaswedQaswed
311313
$endgroup$
In addition to the solution by the OP using the moment generating function, I'll provide a (nearly trivial) solution when the rules about the sum and linear transformations of normal distributions are known.
The distribution of $U-V$ is identical to $U+a cdot V$ with $a=-1$. So from the cited rules we know that $U+Vcdot a sim N(mu_U + acdot mu_V,~sigma_U^2 + a^2 cdot sigma_V^2) = N(mu_U - mu_V,~sigma_U^2 + sigma_V^2)~ text{(for $a = -1$)} = N(0,~2)~text{(for standard normal distributed variables)}$.
Edit 2017-11-20: After I rejected the correction proposed by @Sheljohn of the variance and one typo, several times, he wrote them in a comment, so I finally did see them. Thank you @Sheljohn!
answered Dec 8 '16 at 11:37
QaswedQaswed
311313
edited Nov 20 '17 at 7:30
answered Dec 8 '16 at 11:37
QaswedQaswed
311313
answered Dec 8 '16 at 11:37
QaswedQaswed
311313
answered Dec 8 '16 at 11:37
QaswedQaswed
311313
311313
$begingroup$
If $U$ and $V$ were not independent, would $sigma_{U+V}^2$ be equal to $sigma_U^2+sigma_V^2+2rhosigma_Usigma_V$ where $rho$ is correlation?
$endgroup$
– Milos
Jun 18 '17 at 21:59
2
$begingroup$
@Qaswed -1: $U+aV$ is not distributed as $mathcal{N}( mu_U + amu V, sigma_U^2 + |a| sigma_V^2 )$; $mu_U + amu V$ makes no sense, and the variance is $sigma_U^2 + a^2 sigma_V^2$
$endgroup$
– Sheljohn
Nov 18 '17 at 20:41
2
$begingroup$
@Sheljohn you are right: $a cdot mu V$ is a typo and should be $a cdot mu_V$. And for the variance part it should be $a^2$ instead of $|a|$. I reject the edits as I only thought they are only changes of style. Sorry, my bad!
$endgroup$
– Qaswed
Nov 20 '17 at 7:24
add a comment |
$begingroup$
If $U$ and $V$ were not independent, would $sigma_{U+V}^2$ be equal to $sigma_U^2+sigma_V^2+2rhosigma_Usigma_V$ where $rho$ is correlation?
$endgroup$
– Milos
Jun 18 '17 at 21:59
2
$begingroup$
@Qaswed -1: $U+aV$ is not distributed as $mathcal{N}( mu_U + amu V, sigma_U^2 + |a| sigma_V^2 )$; $mu_U + amu V$ makes no sense, and the variance is $sigma_U^2 + a^2 sigma_V^2$
$endgroup$
– Sheljohn
Nov 18 '17 at 20:41
2
$begingroup$
@Sheljohn you are right: $a cdot mu V$ is a typo and should be $a cdot mu_V$. And for the variance part it should be $a^2$ instead of $|a|$. I reject the edits as I only thought they are only changes of style. Sorry, my bad!
$endgroup$
– Qaswed
Nov 20 '17 at 7:24
$begingroup$
If $U$ and $V$ were not independent, would $sigma_{U+V}^2$ be equal to $sigma_U^2+sigma_V^2+2rhosigma_Usigma_V$ where $rho$ is correlation?
$endgroup$
– Milos
Jun 18 '17 at 21:59
$begingroup$
If $U$ and $V$ were not independent, would $sigma_{U+V}^2$ be equal to $sigma_U^2+sigma_V^2+2rhosigma_Usigma_V$ where $rho$ is correlation?
$endgroup$
– Milos
Jun 18 '17 at 21:59
2
2
$begingroup$
@Qaswed -1: $U+aV$ is not distributed as $mathcal{N}( mu_U + amu V, sigma_U^2 + |a| sigma_V^2 )$; $mu_U + amu V$ makes no sense, and the variance is $sigma_U^2 + a^2 sigma_V^2$
$endgroup$
– Sheljohn
Nov 18 '17 at 20:41
$begingroup$
@Qaswed -1: $U+aV$ is not distributed as $mathcal{N}( mu_U + amu V, sigma_U^2 + |a| sigma_V^2 )$; $mu_U + amu V$ makes no sense, and the variance is $sigma_U^2 + a^2 sigma_V^2$
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– Sheljohn
Nov 18 '17 at 20:41
2
2
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@Sheljohn you are right: $a cdot mu V$ is a typo and should be $a cdot mu_V$. And for the variance part it should be $a^2$ instead of $|a|$. I reject the edits as I only thought they are only changes of style. Sorry, my bad!
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– Qaswed
Nov 20 '17 at 7:24
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@Sheljohn you are right: $a cdot mu V$ is a typo and should be $a cdot mu_V$. And for the variance part it should be $a^2$ instead of $|a|$. I reject the edits as I only thought they are only changes of style. Sorry, my bad!
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– Qaswed
Nov 20 '17 at 7:24
add a comment |
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The currently upvoted answer is wrong, and the author rejected attempts to edit despite 6 reviewers' approval. So here it is; if one knows the rules about the sum and linear transformations of normal distributions, then the distribution of $U-V$ is:
$$
U-V sim U + aV sim mathcal{N}big( mu_U + amu_V, sigma_U^2 + a^2sigma_V^2 big) = mathcal{N}big( mu_U - mu_V, sigma_U^2 + sigma_V^2 big)
$$
where $a=-1$ and $(mu,sigma)$ denote the mean and std for each variable.
answered Nov 18 '17 at 20:44
SheljohnSheljohn
1,291823
$endgroup$
add a comment |
$begingroup$
The currently upvoted answer is wrong, and the author rejected attempts to edit despite 6 reviewers' approval. So here it is; if one knows the rules about the sum and linear transformations of normal distributions, then the distribution of $U-V$ is:
$$
U-V sim U + aV sim mathcal{N}big( mu_U + amu_V, sigma_U^2 + a^2sigma_V^2 big) = mathcal{N}big( mu_U - mu_V, sigma_U^2 + sigma_V^2 big)
$$
where $a=-1$ and $(mu,sigma)$ denote the mean and std for each variable.
answered Nov 18 '17 at 20:44
SheljohnSheljohn
1,291823
$endgroup$
add a comment |
$begingroup$
The currently upvoted answer is wrong, and the author rejected attempts to edit despite 6 reviewers' approval. So here it is; if one knows the rules about the sum and linear transformations of normal distributions, then the distribution of $U-V$ is:
$$
U-V sim U + aV sim mathcal{N}big( mu_U + amu_V, sigma_U^2 + a^2sigma_V^2 big) = mathcal{N}big( mu_U - mu_V, sigma_U^2 + sigma_V^2 big)
$$
where $a=-1$ and $(mu,sigma)$ denote the mean and std for each variable.
answered Nov 18 '17 at 20:44
SheljohnSheljohn
1,291823
$endgroup$
The currently upvoted answer is wrong, and the author rejected attempts to edit despite 6 reviewers' approval. So here it is; if one knows the rules about the sum and linear transformations of normal distributions, then the distribution of $U-V$ is:
$$
U-V sim U + aV sim mathcal{N}big( mu_U + amu_V, sigma_U^2 + a^2sigma_V^2 big) = mathcal{N}big( mu_U - mu_V, sigma_U^2 + sigma_V^2 big)
$$
where $a=-1$ and $(mu,sigma)$ denote the mean and std for each variable.
answered Nov 18 '17 at 20:44
SheljohnSheljohn
1,291823
answered Nov 18 '17 at 20:44
SheljohnSheljohn
1,291823
answered Nov 18 '17 at 20:44
SheljohnSheljohn
1,291823
answered Nov 18 '17 at 20:44
SheljohnSheljohn
1,291823
1,291823
add a comment |
add a comment |
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I think you made a sign error somewhere. The mean of $U-V$ should be zero even if $U$ and $V$ have nonzero mean $mu$. Shouldn't your second line be $E[e^{tU}]E[e^{-tV}]$?
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– Bungo
Sep 2 '14 at 17:54
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Ah, yes it should. Thank you.
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– nonremovable
Sep 2 '14 at 17:55
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Aside from that, your solution looks fine.
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– Bungo
Sep 2 '14 at 17:56
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Yeah, I changed the wrong sign, but in the end the answer still came out to $N(0,2)$.
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– nonremovable
Sep 2 '14 at 17:58
5
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Having $$E[U - V] = E[U] - E[V] = mu_U - mu_V$$ and $$Var(U - V) = Var(U) + Var(V) = sigma_U^2 + sigma_V^2$$ then $$(U - V) sim N(mu_U - mu_V, sigma_U^2 + sigma_V^2)$$
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– Dor
Sep 13 '15 at 8:57