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As title suggest, I am finding the distribution of $(X+a)^2$ where $X\sim N(0,1)$ and $a$ is a constant.

I tried finding it with first principles and transformation formula but the calculation were so messy that I gave up.

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2 Answers 2

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This is the same as $X^2$ where $X$ ~ $N(a,1)$ . Therefore, it has a Noncentral Chi-squared distribution https://en.wikipedia.org/wiki/Noncentral_chi-squared_distribution , with 1 degree of freedom and noncentrality parameter = $a^2$, following the convention of the linked Wikipedia article.

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I tried finding it with first principles and transformation formula but the calculation were so messy that I gave up.

Well, first principles and the transformation formula say that with $Y = (X+a)^2$ and $y > 0$, \begin{align} F_Y(y) &= P\{Y \leq y\}\\ &= P\{(X+a)^2 \leq y\}\\ &= P\{-\sqrt{y} \leq X+a \leq \sqrt{y}\}\\ &= P\{-\sqrt{y}-a \leq X \leq \sqrt{y}-a\}\\ &= F_X(\sqrt{y}-a) - F_X(-\sqrt{y}-a) \end{align} which does not seem to be that hard to do, and if you want the density of $Y$, you can always differentiate with respect to $y$ after re-reading your Calculus I book (especially the part where it says that if the derivative of $G(x)$ is $g(x)$, then the chain rule tells us that the derivative of $G(h(x))$ is $g(h(x))\cdot \frac{\mathrm dh(x)}{\mathrm dx}$), and your probability/statistics/machine-learning book (especially the part which says that the derivative of the CDF $F_X(x)$ is the pdf $f_X(x)$).

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