The following is an exercise problem from Hogg, McKean and Craig's book titled "Introduction to Mathematical Statistics" that I am working on.

3.6.16. Let $X_1$, $X_2$, and $X_3$ be three independent chi-square variables with $r_1$, $r_2$, and $r_3$ degrees of freedom, respectively.

(a) Show that $Y_1 = X_1/X_2$ and $Y_2 = X_1 + X_2$ are independent and that $Y_2$ is $\chi^2(r_1 + r_2)$.

(b) Deduce that $$\cfrac{X_1 /r_1}{X_2/r_2} \ \ \textrm{ and } \ \ \cfrac{X_3 /r_3}{(X_1 + X_2)/(r_1 + r_2)}$$ are independent F-variables.

My attempt:

(a) The inverse of the transformations is given by $X_1 = Y_1Y_2/(1+Y_1)$ and $X_2=Y_2/(1+Y_1)$ and the Jacobian is the determinant of $$ \begin{bmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} \\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} \end{bmatrix} = \begin{bmatrix} y_2/(1+y_1)^2 & y_1/(1+y_1) \\ -y_2/(1+y_1)^2 & 1/(1+y_1) \end{bmatrix} $$ which ends up being $y_2/(1+y_1)^2$. The joint pdf of $Y_1$ and $Y_2$ is then given by

\begin{align}f_{Y_1,Y_2}(y_1,y_2) &= \cfrac{y_2}{(1+y_1)^2} \left( \cfrac{y_1y_2}{1+y_1} \right)^{\left(\frac{r_1}{2}-1\right)}\cfrac{e^{-\left(\frac{y_1y_2}{2(1+y_1)}\right)}}{2^{\frac{r_1}{2}}\Gamma(\frac{r_1}{2})}\left( \cfrac{y_2}{1+y_1} \right)^{\left(\frac{r_2}{2}-1\right)}\cfrac{e^{-\left(\frac{y_2}{2(1+y_1)}\right)}}{2^{\frac{r_2}{2}}\Gamma(\frac{r_2}{2})} \\ &= \cfrac{1}{2^{\left(\frac{r_1+r_2}{2}\right)}\Gamma(\frac{r_1}{2})\Gamma(\frac{r_2}{2})} \times \left[ \frac{y_1^{\left( \frac{r_1}{2}-1\right)}}{(1+y_1)^{\left(\frac{r_1+r_2}{2} \right)}}\right] \times\left[y_2^{\left(\frac{r_1+r_2}{2}-1\right)}e^{-\left(\frac{y_2}{2}\right)}\right]\end{align}.

With that, the pdf nicely splits into a product of a function in $y_1$ alone, and that of $y_2$ alone thereby letting us to conclude that $Y_1$ and $Y_2$ are independent. We need to be careful about the support before we conclude that but as the support is that of positive real numbers for both of them, we should not have a problem on that front.

(b) Let $Y_3 = X_3/(X_1+X_2)$. We need to show that $Y_1$ and $Y_3$ are independent. I will spare the details but taking an approach similar to the one above, we get the Jacobian to be $y_2^2/(1+y_1)^2$ and the joint pdf to be

\begin{align}f_{Y_1,Y_2,Y_3}(y_1,y_2,y_3) &= \cfrac{1}{2^{\left(\frac{r_1+r_2+r_3}{2}\right)}\Gamma(\frac{r_1}{2})\Gamma(\frac{r_2}{2})\Gamma(\frac{r_3}{2})} \times \left[ \frac{y_1^{\left( \frac{r1}{2}-1\right)}y_3^{\left(\frac{r_3}{2}-1\right)}}{(1+y_1)^{\left(\frac{r_1+r_2}{2} \right)}}\right]\times \left[y_2^{\left(\frac{r_1+r_2+r_3}{2}-1\right)}e^{-\left(\frac{y_2(1+y_3)}{2}\right)}\right].\end{align}

Clearly, $Y_1$, $Y_2$ and $Y_3$ are not jointly independent. However, when we integrate out $y_2$ over its support to obtain the marginal pdf $f_{Y_1,Y_3}(y_1,y_3)$, the expression in the second square braces integrates to some kind of Gamma function in $y_3$ along with possibly another factor involving $2$ rasied to a power of $y_3$, which will let us conclude that $Y_1$ and $Y_3$ are independent.

This still lacks details but I think I have explained my main thought process above. Here are my questions to you.

  1. Do you find any error in this thought process?

  2. More importantly, is there a more elegant/sleek solution to part (b) that uses part (a) and some other result? Please note that $Y_1$, $Y_2$, and $Y_3$ are not mutually independent. If they were, part (b) would have been quite trivial.

I am asking this question mainly because in part (b) he starts with "Deduce that ..." by which I feel like there might be a one liner to deduce that $Y_1$ and $Y_3$ are independent. Is there a simpler solution? Please let me know.


2 Answers 2


Your answer to (a) is correct.

For part (b), the two random variables are $F$-distributed by construction. We could prove that they're independent by establishing joint independence of $Y_1, Y_2, X_3$.

I believe you can do this via the joint moment-generating function. We have $$ M_{Y_1,Y_2,X_3}(s,t,u) = \mathbb{E}[\exp(sY_1+tY_2+uX_3)]=\mathbb{E}[\exp(uX_3) \mathbb{E}[\exp(sY_1+tY_2|X_3)] ] $$ We can drop the conditioning on $X_3$ in the inner expectation, as $ (Y_1, Y_2) $ are constructed from $(X_1,X_2)$ only. This gives $$ \mathbb{E}[\exp(sY_1+tY_2|X_3)]=\mathbb{E}[\exp(sY_1+tY_2)]=M_{Y_1,Y_2}(s,t)=M_{Y_1}(s)M_{Y_2}(t), $$ by independence of $Y_1,Y_2$.

Putting it all together, we have $$ M_{Y_1,Y_2,X_3}(s,t,u)=\mathbb{E}[\exp(uX_3) M_{Y_1}(s)M_{Y_2}(t)] =M_{Y_1}(s)M_{Y_2}(t)M_{X_3}(u) $$ and we have joint independence.

  • 1
    $\begingroup$ So even if we grant that $Y_1$, $Y_2$, and $X_3$ are mutually independent, we eventually need to deduce that $Y_1$ and $Y_3 \triangleq X_3/Y_2$ are independent. So does it follow that because $Y_1$, $Y_2$, and $X_3$ are mutually independent, $Y_1$ and $Y_3 \triangleq X_3/Y_2$ are independent? I think so because then $Y_1$ and $f(Y_2,X_3)$ will be independent so we have our result. Is that right? $\endgroup$ Feb 24, 2023 at 12:20
  • 1
    $\begingroup$ Yes, you can deduce that if you have mutual independence, though not pairwise independence (as in my initial, wrong, answer). $\endgroup$ Feb 24, 2023 at 12:22
  • 1
    $\begingroup$ Thanks. So basically I had to look at $Y_1$, $Y_2$ and $X_3$ whereas I was looking at $Y_1$, $Y_2$ and $Y_3$. That's pretty cool! $\endgroup$ Feb 24, 2023 at 12:25
  • 1
    $\begingroup$ It's a nice question, though I wonder if the authors realised that the obvious answer is wrong! It definitely fooled me. $\endgroup$ Feb 24, 2023 at 12:26
  • 1
    $\begingroup$ @TryingHardToBecomeAGoodPrSlvr, I didn't study Hogg&Craig but if I remember, I came across a solution manual or so. $\endgroup$ Feb 24, 2023 at 12:50

To show the independence of $\frac{X_1/r_1}{X_2/r_2}$ and $\frac{X_3/r_3}{(X_1 + X_2)/(r_1 + r_2)}$ in part (b), denote $\frac{X_1/r_1}{X_2/r_2}$ by $Z_1$, $\frac{X_1 + X_2}{r_1 + r_2}$ by $Z_2$, and $\frac{X_3}{r_3}$ by $Z_3$. By the assumption and the result of part (a), we have $Z_1 \perp Z_2$, $Z_3 \perp (Z_1, Z_2)$, whence by Theorem 20.3 in Probability and Measure by Patrick Billingsley, for any $x, y \in \mathbb{R}$: \begin{align} & P\left[\frac{X_1/r_1}{X_2/r_2} \leq x, \frac{X_3/r_3}{(X_1 + X_2)/(r_1 + r_2)} \leq y\right] \\ =& P\left[Z_1 \leq x, \frac{Z_3}{Z_2} \leq y\right] \\ \tag{by $Z_3 \perp (Z_1, Z_2)$} =& \int_0^\infty P[Z_1 \leq x, z Z_2^{-1} \leq y]f_{Z_3}(z)dz \\ =& \int_0^\infty P[Z_1 \leq x]P[z Z_2^{-1} \leq y]f_{Z_3}(z)dz \\ \tag{by $Z_1 \perp Z_2$} =& P[Z_1 \leq x]\int_0^\infty P[z Z_2^{-1} \leq y]f_{Z_3}(z)dz \\ \tag{by $Z_3 \perp Z_2$} =& P[Z_1 \leq x]P\left[\frac{Z_3}{Z_2} \leq y\right] \\ =& P\left[\frac{X_1/r_1}{X_2/r_2} \leq x\right]P\left[\frac{X_3/r_3}{(X_1 + X_2)/(r_1 + r_2)} \leq y\right]. \end{align}

This shows $\frac{X_1/r_1}{X_2/r_2}$ and $\frac{X_3/r_3}{(X_1 + X_2)/(r_1 + r_2)}$ are independent.

Maybe an overkill (but it sheds some geometric insight of this fact), another way to prove part $(a)$ is using spherical coordinates. Since $X_1 \sim \chi_{r_1}^2, X_2 \sim \chi_{r_2}^2$ are independent, we have \begin{align} & X_1 \overset{d}{=} Z_1^2 + \cdots + Z_{r_1}^2, \\ & X_2 \overset{d}{=} Z_{r_1 + 1}^2 + \cdots + Z_{r_1 + r_2}^2, \end{align} where $(Z_1, \ldots, Z_{r_1 + r_2}) \sim N_{r_1 + r_2}(0, I_{(r_1 + r_2)})$.

Express $Z_1, \ldots, Z_{r_1 + r_2}$ in spherical coordinates: \begin{align} & Z_1 = R\cos(\Theta_1) \\ & Z_2 = R\sin(\Theta_1)\cos(\Theta_2) \\ & \cdots\cdots \\ & Z_{r_1 + r_2 - 1} = R\sin(\Theta_1)\cdots\sin(\Theta_{r_1 + r_2 - 2})\cos(\Theta_{r_1 + r_2 - 1}) \\ & Z_{r_1 + r_2} = R\sin(\Theta_1)\cdots\sin(\Theta_{r_1 + r_2 - 2})\sin(\Theta_{r_1 + r_2 - 1}), \end{align} then \begin{align} & \frac{X_1}{X_2} \overset{d}{=} \frac{Z_1^2 + \cdots + Z_{r_1}^2}{Z_{r_1 + 1}^2 + \cdots + Z_{r_1 + r_2}^2} = f(\Theta_1, \ldots, \Theta_{r_1 + r_2 - 1}), \\ & X_1 + X_2 \overset{d}{=} Z_1^2 + \cdots + Z_{r_1 + r_2}^2 = R^2. \tag{1} \end{align}

Since $R$ and $(\Theta_1, \ldots, \Theta_{r_1 + r_2 - 1})$ are independent (because the joint density of them factorizes), it follows by $(1)$ that $\frac{X_1}{X_2}$ and $X_1 + X_2$ are independent.

  • $\begingroup$ Thank you. Also, the thing is that if we know $Z_1=X_1/X_2$, $Z_2=X_1+X_2$, and $X_3$ are mutually independent, then $Z_1$ and $f(Z_2,X_3)$ are independent for any measurable function $f$. I was looking at an inconvenient triplet but as it turns out, the right triplet to consider was $Y_1$, $Y_2$ and $X_3$. That's where I went wrong. :( $\endgroup$ Feb 24, 2023 at 14:48
  • $\begingroup$ With regards to the addition (proof of part (a)), it's a cool thing to express it in terms of spherical co-ordinates. Most of mathematics/problem solving is about seeing "how things move" or "recognizing certain structure" in a given mathematical setup and in this instance, these RVs looked at as co-ordinates helps a lot. Thanks for sharing it with us! $\endgroup$ Feb 25, 2023 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.