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Suppose $X$ is normal with mean $\theta$, and variance equal to 1. What is the distribution of $X^2$?

utobi
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user164144
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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
  • @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
  • 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

1 Answers1

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If $\theta=0$, $X^2$ is a central chi-square with 1 degree of freedom. If $\theta\neq 0$, $X^2$ is a non-central chi-square with 1 degree of freedom and $\theta$ the non-centrality parameter.

utobi
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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