Distribution of $(X+a)^2$ where $X$ is a standard normal and $a$ is a constant 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.
 A: 
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 $f_X(x)$).
A: 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.
