How to Find the CDF and PDF of (X+Y)²?

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SUMMARY

The discussion focuses on finding the cumulative distribution function (CDF) F_Z(z) and probability density function (PDF) f_Z(z) for the random variable Z defined as (X+Y)², where X and Y are independent random variables. The process involves two main steps: first, determining the distribution function for W = X + Y through convolution of the individual distribution functions F_X(x) and F_Y(y). Next, the distribution function for Z is derived by evaluating P(Z < z) as P(-√z < W < √z).

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EngWiPy
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Hi,

Suppose we have two random variables [tex]X[/tex] and [tex]Y[/tex] with CDFs and PDFs [tex]F_X(x)[/tex], [tex]F_Y(y)[/tex], [tex]f_X(x)[/tex], and [tex]f_Y(y)[/tex]. Now suppose that a new random variable formed as [tex]Z=(X+Y)^2[/tex]. How can we find the CDF [tex]F_Z(z)[/tex] and the PDF [tex]f_Z(z)[/tex] of this new random variable? Note: [tex]X[/tex] and [tex]Y[/tex] are independent random variables.

Thanks
 
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If you do it in two steps it is straightforward.
First get the distribution function for W=X+Y (convolution of the distribution functions).
Then get the distribution function for the square of a random variable Z = W2.
P(Z < z)=P(-√z < W < √z).
 
mathman said:
If you do it in two steps it is straightforward.
First get the distribution function for W=X+Y (convolution of the distribution functions).
Then get the distribution function for the square of a random variable Z = W2.
P(Z < z)=P(-√z < W < √z).

I though in it in another way:

[tex]W_1=X^2+Y^2[/tex] and [tex]W_2=2XY[/tex], then [tex]Z=W_1+W_2[/tex]. Your approach seems easier.
 

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