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Find the density function of $Z=X+Y$ when $X$ and $Y$ are standard normal [duplicate]
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Proof that sum of independent normals is normal using convolutions
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Two independent random variables $X$ and $Y$ have standard normal distributions. Find the density function of random variable $Z=X+Y$ using the convolution method.
This shows this problem on page 4 but it seems to use some tricks that are, well, pretty tricky. In particular getting behind their third line is beyond me. Is there an alternate way to show this that is more straightforward perhaps even with more (unskipped) steps?
probability random-variables convolution
asked Jul 17 at 21:21
Frank Shmrank
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marked as duplicate by amWhy, heropup probability
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Jul 17 at 22:07
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This question already has an answer here:
Proof that sum of independent normals is normal using convolutions
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Two independent random variables $X$ and $Y$ have standard normal distributions. Find the density function of random variable $Z=X+Y$ using the convolution method.
This shows this problem on page 4 but it seems to use some tricks that are, well, pretty tricky. In particular getting behind their third line is beyond me. Is there an alternate way to show this that is more straightforward perhaps even with more (unskipped) steps?
probability random-variables convolution
asked Jul 17 at 21:21
Frank Shmrank
279112
marked as duplicate by amWhy, heropup probability
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This question already has an answer here:
Proof that sum of independent normals is normal using convolutions
1 answer
Two independent random variables $X$ and $Y$ have standard normal distributions. Find the density function of random variable $Z=X+Y$ using the convolution method.
This shows this problem on page 4 but it seems to use some tricks that are, well, pretty tricky. In particular getting behind their third line is beyond me. Is there an alternate way to show this that is more straightforward perhaps even with more (unskipped) steps?
probability random-variables convolution
asked Jul 17 at 21:21
Frank Shmrank
279112
This question already has an answer here:
Proof that sum of independent normals is normal using convolutions
1 answer
Two independent random variables $X$ and $Y$ have standard normal distributions. Find the density function of random variable $Z=X+Y$ using the convolution method.
This shows this problem on page 4 but it seems to use some tricks that are, well, pretty tricky. In particular getting behind their third line is beyond me. Is there an alternate way to show this that is more straightforward perhaps even with more (unskipped) steps?
This question already has an answer here:
Proof that sum of independent normals is normal using convolutions
1 answer
probability random-variables convolution
asked Jul 17 at 21:21
Frank Shmrank
279112
edited Jul 17 at 22:10
user401938
asked Jul 17 at 21:21
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asked Jul 17 at 21:21
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asked Jul 17 at 21:21
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Jul 17 at 22:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
$$f_Z(z) = int_-infty^infty f_X(z - y)f_Y(y)dy \=
frac12 piint_-infty^infty e^-frac(z - y)^22e^-fracy^22dy \= frac12 piint_-infty^infty expleft(-fracz^2 - 2zy + 2y^22right)dy \= frac12 piint_-infty^infty expleft(-fracfracz^22 - 2zy + 2y^2 + fracz^222right)dy \= frac12 piint_-infty^infty expleft(-frac(fraczsqrt2 - sqrt2y)^2 + fracz^222right)dy \= frace^-fracz^242 piint_-infty^infty expleft(-(fraczsqrt2 - sqrt2y)^2 / 2right)dy \= frace^-fracz^242pi times sqrtpi \= frace^frac-z^24sqrt4 pi$$
So, $Z sim N(0, 2)$
edited Jul 19 at 19:53
Frank Shmrank
279112
answered Jul 17 at 21:39
D F
1,0551218
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A very simple way of finding the density of Z is to recognize that the sum of Gaussians is still Gaussian. Then you can identify the distribution by finding the mean and variance of Z.
$$mu _Z = E_Zleft[ Z right] = E_Xleft[ X right] + E_Yleft[ Y right] = mu _X + mu _Y = 0$$
and
$$beginarrayl
sigma _Z^2 = E_Zleft[ Z^2 right] - mu _Z^2 = E_Xleft[ X^2 right] + 2E_Xleft[ X right]E_Yleft[ Y right] + E_Yleft[ Y^2 right] - mu _Z^2\
= left( sigma _X^2 + mu _X^2 right) + 2mu _Xmu _Y + left( sigma _Y^2 + mu _Y^2 right) - left( mu _X^2 + 2mu _Xmu _Y + mu _Y^2 right) = sigma _X^2 + sigma _Y^2 = 2
endarray$$
answered Jul 17 at 21:43
John Polcari
382111
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$f_Z(z) = int_-infty^infty f_X(z - y)f_Y(y)dy \=
frac12 piint_-infty^infty e^-frac(z - y)^22e^-fracy^22dy \= frac12 piint_-infty^infty expleft(-fracz^2 - 2zy + 2y^22right)dy \= frac12 piint_-infty^infty expleft(-fracfracz^22 - 2zy + 2y^2 + fracz^222right)dy \= frac12 piint_-infty^infty expleft(-frac(fraczsqrt2 - sqrt2y)^2 + fracz^222right)dy \= frace^-fracz^242 piint_-infty^infty expleft(-(fraczsqrt2 - sqrt2y)^2 / 2right)dy \= frace^-fracz^242pi times sqrtpi \= frace^frac-z^24sqrt4 pi$$
So, $Z sim N(0, 2)$
edited Jul 19 at 19:53
Frank Shmrank
279112
answered Jul 17 at 21:39
D F
1,0551218
add a comment |Â
up vote
3
down vote
accepted
$$f_Z(z) = int_-infty^infty f_X(z - y)f_Y(y)dy \=
frac12 piint_-infty^infty e^-frac(z - y)^22e^-fracy^22dy \= frac12 piint_-infty^infty expleft(-fracz^2 - 2zy + 2y^22right)dy \= frac12 piint_-infty^infty expleft(-fracfracz^22 - 2zy + 2y^2 + fracz^222right)dy \= frac12 piint_-infty^infty expleft(-frac(fraczsqrt2 - sqrt2y)^2 + fracz^222right)dy \= frace^-fracz^242 piint_-infty^infty expleft(-(fraczsqrt2 - sqrt2y)^2 / 2right)dy \= frace^-fracz^242pi times sqrtpi \= frace^frac-z^24sqrt4 pi$$
So, $Z sim N(0, 2)$
edited Jul 19 at 19:53
Frank Shmrank
279112
answered Jul 17 at 21:39
D F
1,0551218
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$f_Z(z) = int_-infty^infty f_X(z - y)f_Y(y)dy \=
frac12 piint_-infty^infty e^-frac(z - y)^22e^-fracy^22dy \= frac12 piint_-infty^infty expleft(-fracz^2 - 2zy + 2y^22right)dy \= frac12 piint_-infty^infty expleft(-fracfracz^22 - 2zy + 2y^2 + fracz^222right)dy \= frac12 piint_-infty^infty expleft(-frac(fraczsqrt2 - sqrt2y)^2 + fracz^222right)dy \= frace^-fracz^242 piint_-infty^infty expleft(-(fraczsqrt2 - sqrt2y)^2 / 2right)dy \= frace^-fracz^242pi times sqrtpi \= frace^frac-z^24sqrt4 pi$$
So, $Z sim N(0, 2)$
edited Jul 19 at 19:53
Frank Shmrank
279112
answered Jul 17 at 21:39
D F
1,0551218
$$f_Z(z) = int_-infty^infty f_X(z - y)f_Y(y)dy \=
frac12 piint_-infty^infty e^-frac(z - y)^22e^-fracy^22dy \= frac12 piint_-infty^infty expleft(-fracz^2 - 2zy + 2y^22right)dy \= frac12 piint_-infty^infty expleft(-fracfracz^22 - 2zy + 2y^2 + fracz^222right)dy \= frac12 piint_-infty^infty expleft(-frac(fraczsqrt2 - sqrt2y)^2 + fracz^222right)dy \= frace^-fracz^242 piint_-infty^infty expleft(-(fraczsqrt2 - sqrt2y)^2 / 2right)dy \= frace^-fracz^242pi times sqrtpi \= frace^frac-z^24sqrt4 pi$$
So, $Z sim N(0, 2)$
edited Jul 19 at 19:53
Frank Shmrank
279112
answered Jul 17 at 21:39
D F
1,0551218
edited Jul 19 at 19:53
Frank Shmrank
279112
edited Jul 19 at 19:53
Frank Shmrank
279112
edited Jul 19 at 19:53
Frank Shmrank
279112
279112
answered Jul 17 at 21:39
D F
1,0551218
answered Jul 17 at 21:39
D F
1,0551218
answered Jul 17 at 21:39
D F
1,0551218
1,0551218
add a comment |Â
add a comment |Â
up vote
3
down vote
A very simple way of finding the density of Z is to recognize that the sum of Gaussians is still Gaussian. Then you can identify the distribution by finding the mean and variance of Z.
$$mu _Z = E_Zleft[ Z right] = E_Xleft[ X right] + E_Yleft[ Y right] = mu _X + mu _Y = 0$$
and
$$beginarrayl
sigma _Z^2 = E_Zleft[ Z^2 right] - mu _Z^2 = E_Xleft[ X^2 right] + 2E_Xleft[ X right]E_Yleft[ Y right] + E_Yleft[ Y^2 right] - mu _Z^2\
= left( sigma _X^2 + mu _X^2 right) + 2mu _Xmu _Y + left( sigma _Y^2 + mu _Y^2 right) - left( mu _X^2 + 2mu _Xmu _Y + mu _Y^2 right) = sigma _X^2 + sigma _Y^2 = 2
endarray$$
answered Jul 17 at 21:43
John Polcari
382111
add a comment |Â
up vote
3
down vote
A very simple way of finding the density of Z is to recognize that the sum of Gaussians is still Gaussian. Then you can identify the distribution by finding the mean and variance of Z.
$$mu _Z = E_Zleft[ Z right] = E_Xleft[ X right] + E_Yleft[ Y right] = mu _X + mu _Y = 0$$
and
$$beginarrayl
sigma _Z^2 = E_Zleft[ Z^2 right] - mu _Z^2 = E_Xleft[ X^2 right] + 2E_Xleft[ X right]E_Yleft[ Y right] + E_Yleft[ Y^2 right] - mu _Z^2\
= left( sigma _X^2 + mu _X^2 right) + 2mu _Xmu _Y + left( sigma _Y^2 + mu _Y^2 right) - left( mu _X^2 + 2mu _Xmu _Y + mu _Y^2 right) = sigma _X^2 + sigma _Y^2 = 2
endarray$$
answered Jul 17 at 21:43
John Polcari
382111
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A very simple way of finding the density of Z is to recognize that the sum of Gaussians is still Gaussian. Then you can identify the distribution by finding the mean and variance of Z.
$$mu _Z = E_Zleft[ Z right] = E_Xleft[ X right] + E_Yleft[ Y right] = mu _X + mu _Y = 0$$
and
$$beginarrayl
sigma _Z^2 = E_Zleft[ Z^2 right] - mu _Z^2 = E_Xleft[ X^2 right] + 2E_Xleft[ X right]E_Yleft[ Y right] + E_Yleft[ Y^2 right] - mu _Z^2\
= left( sigma _X^2 + mu _X^2 right) + 2mu _Xmu _Y + left( sigma _Y^2 + mu _Y^2 right) - left( mu _X^2 + 2mu _Xmu _Y + mu _Y^2 right) = sigma _X^2 + sigma _Y^2 = 2
endarray$$
answered Jul 17 at 21:43
John Polcari
382111
A very simple way of finding the density of Z is to recognize that the sum of Gaussians is still Gaussian. Then you can identify the distribution by finding the mean and variance of Z.
$$mu _Z = E_Zleft[ Z right] = E_Xleft[ X right] + E_Yleft[ Y right] = mu _X + mu _Y = 0$$
and
$$beginarrayl
sigma _Z^2 = E_Zleft[ Z^2 right] - mu _Z^2 = E_Xleft[ X^2 right] + 2E_Xleft[ X right]E_Yleft[ Y right] + E_Yleft[ Y^2 right] - mu _Z^2\
= left( sigma _X^2 + mu _X^2 right) + 2mu _Xmu _Y + left( sigma _Y^2 + mu _Y^2 right) - left( mu _X^2 + 2mu _Xmu _Y + mu _Y^2 right) = sigma _X^2 + sigma _Y^2 = 2
endarray$$
answered Jul 17 at 21:43
John Polcari
382111
answered Jul 17 at 21:43
John Polcari
382111
answered Jul 17 at 21:43
John Polcari
382111
answered Jul 17 at 21:43
John Polcari
382111
382111
add a comment |Â
add a comment |Â