Say we have to independent normal distributions ($X$ and $Y$). What is the distribution of Z where Z is $(X + Y)^2$.


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    $\begingroup$ Can you figure out what the distribution of $X+Y$ is? Once you know that, figuring out the distribution of $Z = (X+Y)^2$ is a lot easier. Most textbooks give general formulas for the squaring transformation which can be applied to this case. $\endgroup$ Sep 19, 2018 at 2:17
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    $\begingroup$ Since $-Y$ also has a Normal distribution, $Z=X-(-Y)$ is the difference of Normal distributions, so you can read the answer at stats.stackexchange.com/questions/186463/…. $\endgroup$
    – whuber
    Sep 19, 2018 at 14:59

1 Answer 1


Your random variable $Z$ has a distribution related to the non-central chi-squared distribution.

Here is a specific example:

Suppose $X \sim \mathsf{Norm}(20, 3)$ and independently $X \sim \mathsf{Norm}(30, 4).$ Then $T = X + Y \sim \mathsf{Norm}(50, \sqrt{3^2+4^2}=5)$ and $T/5 \sim \mathsf{Norm}(10,1).$

Then $Z = T^2/25 \sim \mathsf{Chisq}(df = 1, ncp=10^2),$ a non-contral chi-squared distribution with noncentrality parameter 100, according to the definition of that distribution given in the Wikipedia link above.

By way of demonstration, consider the brief simulation in R below.

t = rnorm(10^5, 50, 5);  z = t^2/25
hist(z, prob=T, br=30, col="skyblue2", main="Sample from CHISQ(df=1, ncp=100)")
curve(dchisq(x, 1, ncp=100), add=T, col="red", lwd=2)

enter image description here

Note: Non-central chi-squared distributions are widely used in applications to model the sum of squares of several independent normal random variables with various means and common unit standard deviation. One specific application is in finding the power of a one-factor, fixed-effect ANOVA.


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