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Let $X_1$ and $X_2$ be independent normally distributed random variables.

$$X_1 \sim N(0,1) \\ X_2 \sim N(0,1)$$ Find the pdf of $\frac{(X_1-X_2)^2}{2}$

We have that

By example we know that when $Y = X^2$ we have that

$$\begin{align}f_Y(y) &= \frac{e^{-\frac{(-\sqrt{y})^2}{2}}}{\sqrt{2\pi}}\left|-\frac{1}{2\sqrt{y}}\right|+\frac{e^{-\frac{(\sqrt{y})^2}{2}}}{\sqrt{2\pi}}\left|\frac{1}{2\sqrt{y}}\right| \\ &= \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{y}}e^{-\frac{y}{2}} \end{align}$$

Which follows achi-square distribution with 1 degree of freedom, therefore

$$\frac{(X_1-X_2)^2}{2} \sim \chi_1^2$$

However, how can I do the calculation for this using multivariate transformations, I cannot get the distribution?

dollar bill
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    Note for $y\geq 0$ fixed, $$\mathbb{P}\left(\frac{(X_1-X_2)^2}{2} \leq y\right)=\mathbb{P}\left(-\sqrt{2y} \leq X_1-X_2 \leq \sqrt{2y}\right)$$ where $X_1-X_2\sim\mathcal{N}(0,2)$. Can you finish? – Matthew H. Apr 02 '22 at 01:09
  • @MatthewH. I know we can have the following $f_{Y}(y) = \frac{1}{2\sqrt{y}}(f_X(\sqrt{y}) + f_X(-\sqrt{y}))$ and so $f_X(x) = \frac{e^{-\frac{x^2}{2}}}{\sqrt{2 \pi}}$. I know how to do this for a single random variable, however when $X_1, X_2$ are involved I'm not sure. For example, $F_Y(y) = P(-\sqrt{2y}) < X_1 - X_2 < \sqrt{2y} = P(X_1 - X_2 \le \sqrt{2y}) - P(X_1-X_2 \le -\sqrt{2y})$ – dollar bill Apr 05 '22 at 15:14

1 Answers1

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Let $U=\frac{(X_1-X_2)}{\sqrt{2}}$ and $M_{x_i(t)}$ be moment generating function of $X_i$.
$M_{\frac{X_1-X_2}{\sqrt{2}}}(t)=E(e^{(\frac{X_1-X_2}{\sqrt{2}})t})=E(e^{\frac{X_1}{\sqrt{2}}t}) E(e^{\frac{X_1}{\sqrt{2}}t})=M_{X_1}(\frac{t}{\sqrt{2}})M_{X_2}(\frac{-t}{\sqrt{2}})=e^{\frac{1}{2}(\frac{t}{\sqrt{2}})^2}e^{\frac{1}{2}(\frac{-t}{\sqrt{2}})^2}=e^{\frac{t^2}{2}}$
So $U=\frac{(X_1-X_2)}{\sqrt{2}} \sim N(0,1)$ Therefore $U^2=(\frac{X_1-X_2}{\sqrt{2}})^2 \sim \chi^2_1$

saleh din
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