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

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To find the CDF F_Z(z) and PDF f_Z(z) of the random variable Z=(X+Y)², start by determining the distribution function for W=X+Y using the convolution of the individual CDFs F_X(x) and F_Y(y). Once W is established, the next step is to derive the distribution function for Z by calculating P(Z < z), which can be expressed as P(-√z < W < √z). An alternative approach involves defining W_1=X²+Y² and W_2=2XY, leading to Z=W_1+W_2, but the convolution method is deemed simpler. This two-step process effectively simplifies the calculation of the CDF and PDF for the squared sum of independent random variables.
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Hi,

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

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