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

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 $$W_2$$?

• It would be helpful to edit the title to at least contain the topic of the question. This will attract better answers probably. Feb 22, 2021 at 1:30
• $X_1 - X_2$ is zero mean, it is just that the variance is not necessarily 1. Thus the square has distribution proportional to a chi-square.
– mjw
Feb 22, 2021 at 1:40
• The $2\sigma^2$ is the factor that would transform your difference into a standard normal and is thus exactly what to multiply a chi-square by to get the distribution of $(X_1-X_2)^2$.
– mjw
Feb 22, 2021 at 1:42
• Ohhh I see it now. The 0 mean threw me off. Ok, thank you so much! Feb 22, 2021 at 1:44
• Note that F(1,1) does not have a mean, thus also no variance. Feb 22, 2021 at 1:49

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*}