If $E \left(\frac{X}{Y}\right)=3$, and $m+n=14$. Then $E \left(\frac{Y}{X}\right)$?

Question

Let $X$ and $Y$ are independently distributed central chi-squad random variables with degrees of freedom $m(\geq 3)$ and $n(\geq 3)$, respectively. If $E \left(\frac{X}{Y}\right)=3$, and $m+n=14$. Then $E \left(\frac{Y}{X}\right)$?

Answer 1

It is a well known result that for independent chi-squared distributions, $X,Y$ with $m, n$ degrees of freedom respectively, then:

$$ Z = \frac{X/m}{Y/n} \sim F(m, n) $$

(follows an F distribution with $m,n$ degrees of freedom). A google search will yield a proof of this.

Now, using the fact that the mean of an $F(m, n)$ distribution, which is only defined for $n \ge 2$, is $\frac{n}{n-2}$, we have:

\begin{align*} E \left ( \frac{X}{Y} \right) &= \frac{m}{n}E \left ( \frac{X}{Y} \frac{n}{m} \right)\\ & = \frac{m}{n} E \left( \frac{X/m}{Y/n} \right )\\ & = \frac{m}{n} E(Z)\\ & = \frac{m}{n} \frac{n}{n-2}\\ & = \frac{m}{n-2} \end{align*}

Using the first piece of information that $E \left ( \frac{X}{Y} \right) = 3$ gives:

$$ m = 3(n-2) $$

We also are given that $m+n=14$ and so $m = 14-n$. Solving this system of $2$ equations:

$$ 3n - 6 = 14 - n \implies n = 5, m = 9 $$

Now, we have that $X = \chi^2_9, Y =\chi^2_5 $.

Therefore:

\begin{align*} E \left ( \frac{Y}{X} \right) &= \frac{n}{m} E \left ( \frac{Y}{X} \frac{m}{n} \right ) \\ & = \frac{n}{m} E \left ( \frac{Y/n}{X/m} \right ) \\ & = \frac{n}{m} E(\tilde{Z}) \end{align*}

where $\tilde{Z} = F(n, m)$ and $E(\tilde{Z}) = \frac{m}{m-2}$.

and so:

$$ E \left ( \frac{Y}{X} \right) = \frac{n}{m} E(\tilde{Z}) = \frac{n}{m} \frac{m}{m-2} = \frac{n}{m-2} =\frac{5}{7} $$