Derivation of sum of Normal RVs PDF
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$begingroup$
Let X~N(0,1) and Y~N(0,1) and independent. Find PDF $f_{x+y}(w)$ by convolution where w=x+y.
So far, this is what I have...
$int_{-infty}^{infty}f_{x}f_{y}(w-x)dx$
$int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(x^2)] * frac{1}{sqrt{2pi}}Exp[frac{-1}{2}((w-x)^2)] dx$
$int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(x^2)]*frac{1}{sqrt{2pi}}Exp[frac{-1}{2}((w^2-2xw+x^2)] dx$
$frac{1}{sqrt{2pi}}Exp[frac{-1}{2}w^2]*int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(-2xw+x^2)dx$]
Not sure where to go from here...I know ultimately the PDF will follow N(0,2)..Should my limits be from 0 to w, or negative infinity to infinity?
Thank you.
probability-distributions normal-distribution convolution
asked Dec 17 '18 at 4:25
user627099user627099
11
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add a comment |
$begingroup$
Let X~N(0,1) and Y~N(0,1) and independent. Find PDF $f_{x+y}(w)$ by convolution where w=x+y.
So far, this is what I have...
$int_{-infty}^{infty}f_{x}f_{y}(w-x)dx$
$int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(x^2)] * frac{1}{sqrt{2pi}}Exp[frac{-1}{2}((w-x)^2)] dx$
$int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(x^2)]*frac{1}{sqrt{2pi}}Exp[frac{-1}{2}((w^2-2xw+x^2)] dx$
$frac{1}{sqrt{2pi}}Exp[frac{-1}{2}w^2]*int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(-2xw+x^2)dx$]
Not sure where to go from here...I know ultimately the PDF will follow N(0,2)..Should my limits be from 0 to w, or negative infinity to infinity?
Thank you.
probability-distributions normal-distribution convolution
asked Dec 17 '18 at 4:25
user627099user627099
11
$endgroup$
add a comment |
$begingroup$
Let X~N(0,1) and Y~N(0,1) and independent. Find PDF $f_{x+y}(w)$ by convolution where w=x+y.
So far, this is what I have...
$int_{-infty}^{infty}f_{x}f_{y}(w-x)dx$
$int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(x^2)] * frac{1}{sqrt{2pi}}Exp[frac{-1}{2}((w-x)^2)] dx$
$int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(x^2)]*frac{1}{sqrt{2pi}}Exp[frac{-1}{2}((w^2-2xw+x^2)] dx$
$frac{1}{sqrt{2pi}}Exp[frac{-1}{2}w^2]*int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(-2xw+x^2)dx$]
Not sure where to go from here...I know ultimately the PDF will follow N(0,2)..Should my limits be from 0 to w, or negative infinity to infinity?
Thank you.
probability-distributions normal-distribution convolution
asked Dec 17 '18 at 4:25
user627099user627099
11
$endgroup$
Let X~N(0,1) and Y~N(0,1) and independent. Find PDF $f_{x+y}(w)$ by convolution where w=x+y.
So far, this is what I have...
$int_{-infty}^{infty}f_{x}f_{y}(w-x)dx$
$int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(x^2)] * frac{1}{sqrt{2pi}}Exp[frac{-1}{2}((w-x)^2)] dx$
$int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(x^2)]*frac{1}{sqrt{2pi}}Exp[frac{-1}{2}((w^2-2xw+x^2)] dx$
$frac{1}{sqrt{2pi}}Exp[frac{-1}{2}w^2]*int_{0}^{w}frac{1}{sqrt{2pi}}Exp[frac{-1}{2}(-2xw+x^2)dx$]
Not sure where to go from here...I know ultimately the PDF will follow N(0,2)..Should my limits be from 0 to w, or negative infinity to infinity?
Thank you.
probability-distributions normal-distribution convolution
probability-distributions normal-distribution convolution
asked Dec 17 '18 at 4:25
user627099user627099
11
asked Dec 17 '18 at 4:25
user627099user627099
11
asked Dec 17 '18 at 4:25
user627099user627099
11
asked Dec 17 '18 at 4:25
user627099user627099
11
asked Dec 17 '18 at 4:25
user627099user627099
11
11
add a comment |
add a comment |
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