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


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 $W_2$?

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

1 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*}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .