Two stability of normal random variable detailed explanation
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I am reading about two stability of normal random variable and I am super confused (no good background in stats or calc), but I really need to understand it. Please, see the following:
I first got stuck on "By definition, $f_z(z) = int f_x(z-y) f_y(y) dy$"
Could someone explain me how we get $f_z$ ? I do not really understand the intuition behind the integral. I am further confused about the computation. It would be great if you could show me intermediate steps of this computation (integral) so I can follow. I will really appreciate!
calculus probability statistics
asked Dec 16 '18 at 1:54
YohanRothYohanRoth
6381714
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add a comment |
$begingroup$
I am reading about two stability of normal random variable and I am super confused (no good background in stats or calc), but I really need to understand it. Please, see the following:
I first got stuck on "By definition, $f_z(z) = int f_x(z-y) f_y(y) dy$"
Could someone explain me how we get $f_z$ ? I do not really understand the intuition behind the integral. I am further confused about the computation. It would be great if you could show me intermediate steps of this computation (integral) so I can follow. I will really appreciate!
calculus probability statistics
asked Dec 16 '18 at 1:54
YohanRothYohanRoth
6381714
$endgroup$
$begingroup$
It is a convolution (the keyword might help): summing independent random variables corresponds to doing a convolution of their probability density functions.
$endgroup$
– Clement C.
Dec 16 '18 at 2:03
1
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I'm a little disturbed that the "claim" fails to mention that $X$ and $Y$ are independent.
$endgroup$
– zoidberg
Dec 16 '18 at 2:48
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@ClementC. thanks for the reference!
$endgroup$
– YohanRoth
Dec 16 '18 at 4:21
add a comment |
$begingroup$
I am reading about two stability of normal random variable and I am super confused (no good background in stats or calc), but I really need to understand it. Please, see the following:
I first got stuck on "By definition, $f_z(z) = int f_x(z-y) f_y(y) dy$"
Could someone explain me how we get $f_z$ ? I do not really understand the intuition behind the integral. I am further confused about the computation. It would be great if you could show me intermediate steps of this computation (integral) so I can follow. I will really appreciate!
calculus probability statistics
asked Dec 16 '18 at 1:54
YohanRothYohanRoth
6381714
$endgroup$
I am reading about two stability of normal random variable and I am super confused (no good background in stats or calc), but I really need to understand it. Please, see the following:
I first got stuck on "By definition, $f_z(z) = int f_x(z-y) f_y(y) dy$"
Could someone explain me how we get $f_z$ ? I do not really understand the intuition behind the integral. I am further confused about the computation. It would be great if you could show me intermediate steps of this computation (integral) so I can follow. I will really appreciate!
calculus probability statistics
calculus probability statistics
asked Dec 16 '18 at 1:54
YohanRothYohanRoth
6381714
asked Dec 16 '18 at 1:54
YohanRothYohanRoth
6381714
asked Dec 16 '18 at 1:54
YohanRothYohanRoth
6381714
asked Dec 16 '18 at 1:54
YohanRothYohanRoth
6381714
asked Dec 16 '18 at 1:54
YohanRothYohanRoth
6381714
6381714
$begingroup$
It is a convolution (the keyword might help): summing independent random variables corresponds to doing a convolution of their probability density functions.
$endgroup$
– Clement C.
Dec 16 '18 at 2:03
1
$begingroup$
I'm a little disturbed that the "claim" fails to mention that $X$ and $Y$ are independent.
$endgroup$
– zoidberg
Dec 16 '18 at 2:48
$begingroup$
@ClementC. thanks for the reference!
$endgroup$
– YohanRoth
Dec 16 '18 at 4:21
add a comment |
$begingroup$
It is a convolution (the keyword might help): summing independent random variables corresponds to doing a convolution of their probability density functions.
$endgroup$
– Clement C.
Dec 16 '18 at 2:03
1
$begingroup$
I'm a little disturbed that the "claim" fails to mention that $X$ and $Y$ are independent.
$endgroup$
– zoidberg
Dec 16 '18 at 2:48
$begingroup$
@ClementC. thanks for the reference!
$endgroup$
– YohanRoth
Dec 16 '18 at 4:21
$begingroup$
It is a convolution (the keyword might help): summing independent random variables corresponds to doing a convolution of their probability density functions.
$endgroup$
– Clement C.
Dec 16 '18 at 2:03
$begingroup$
It is a convolution (the keyword might help): summing independent random variables corresponds to doing a convolution of their probability density functions.
$endgroup$
– Clement C.
Dec 16 '18 at 2:03
1
1
$begingroup$
I'm a little disturbed that the "claim" fails to mention that $X$ and $Y$ are independent.
$endgroup$
– zoidberg
Dec 16 '18 at 2:48
$begingroup$
I'm a little disturbed that the "claim" fails to mention that $X$ and $Y$ are independent.
$endgroup$
– zoidberg
Dec 16 '18 at 2:48
$begingroup$
@ClementC. thanks for the reference!
$endgroup$
– YohanRoth
Dec 16 '18 at 4:21
$begingroup$
@ClementC. thanks for the reference!
$endgroup$
– YohanRoth
Dec 16 '18 at 4:21
add a comment |
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$begingroup$
It is a convolution (the keyword might help): summing independent random variables corresponds to doing a convolution of their probability density functions.
$endgroup$
– Clement C.
Dec 16 '18 at 2:03
1
$begingroup$
I'm a little disturbed that the "claim" fails to mention that $X$ and $Y$ are independent.
$endgroup$
– zoidberg
Dec 16 '18 at 2:48
$begingroup$
@ClementC. thanks for the reference!
$endgroup$
– YohanRoth
Dec 16 '18 at 4:21