F-distributions and chi-square distributions: Can someone help explain the answer to this problem?

Question

Let $X_1, X_2, X_3, X_4$ be independent $\mathscr{N}(\mu, \sigma^2)$ random variables. Find the distribution of the ratio

$$R=\frac{(X_1-X_2)^2}{(X_3-X_4)^2}.$$

Note that $X_1-X_2$ and $X_3-X_4$ are independent $\mathscr{N}(0, \sigma^2)$ random variables. Thus $W_1=(X_1-X_2)^2/(2\sigma^2)$ and $W_2=(X_3-X_4)^2/(2\sigma^2)$ are independent $\mathcal{x}^2(1)$ random variables. Since $R$ equals $W_1/W_2$, we conclude $R\sim F(1,1).$

I understand that an $F$ distribution is the ratio of two chi-squares with their respective degrees of freedom. But for this one, why is the square of a non-standard normal distribution a chi-square? Why is it only one degree of freedom? Moreover, where does the $2\sigma^2$ come from on the denominators for $W_1$ and $W_2$?

Answer 1

Note that $Y_1=X_1-X_2\sim N(0,2\sigma ^2)$ and $Y_2=X_3-X_4\sim N(0,2\sigma ^2)$. We can then normalize these into standard normal random variables: \begin{align*} Z_1=\frac{X_1-X_2}{2\sigma^2}\sim N(0,1)\\ Z_2=\frac{X_3-X_4}{2\sigma^2}\sim N(0,1) \end{align*} Since $Z_1,Z_2$ are standard normal, we can say that $$\frac{Z_1^2}{Z_2^2}\sim F(1,1)$$ So \begin{align*} \frac{\left(\frac{X_1-X_2}{2\sigma^2}\right)^2}{\left(\frac{X_3-X_4}{2\sigma^2}\right)^2}\sim F(1,1)\\ \frac{(X_1-X_2)^2}{(X_3-X_4)^2}\sim F(1,1) \end{align*}