Suppose X is normal ($\theta$, 1). What is the distribution of X$^2$?
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Why are short concise questions so often considered to be low quality? If it is a duplicate close it for that reason. – Michael R. Chernick Jun 07 '17 at 23:36
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@Michael, if it's a duplicate (which this certainly is, many times over), *please close it as one*. Search for a duplicate (which may be more convenient done in another tab), click close on this, click through to closing as duplicate, paste in the url. If you make a reasonable attempt to find one and can't, then sure, answer it (unless it looks like homework...) -- but a duplicate of this one could be found by basic searching (like *square of a normal*) – Glen_b Jun 08 '17 at 01:06
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I am not surprised. I was just responding to it being in the queue as low quality. I am not surprised that you found other questions that are duplicates. I just left it for someone else to search. – Michael R. Chernick Jun 08 '17 at 01:16
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If $\theta$=0, $X^2$ is a central chi square with 1 degree of freedom. If $\theta$ is not equal to 0 $X^2$ is a non-central chi square with 1 degree of freedom and $\theta$ the non-centrality parameter.
Michael R. Chernick
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Sometimes as described in wikipedia $\theta$$^2$ is referred to as the non-centrality parameter. This generalizes when you sum the squares of several i.i.d. normal random variables with variance 1. – Michael R. Chernick Jun 07 '17 at 23:28