# Expected value of $X^{-1}$, $X$ being a noncentral $\chi^2$. Cannot understand a step of a equation in a paper

I am reading the following paper:

Mudholkar GS, Chaubey YP, Ching-Chuong L (1976). Approximations for the doubly noncentral-$F$ distribution. Communications in Statistics - Theory and Methods, 5(1):49–63. doi:10.1080/03610927608827331

In this paper (section 2, page 51), $X$ and $Y$ are both defined to be noncentral $\chi^2$ random variables, with respective degrees of freedom $\nu_1$ and $\nu_2$ and noncentrality parameters $\lambda_1$ and $\lambda_2$. $X$ and $Y$ are also assumed to be independent.

The authors try to obtain an approximation to the raw moments of the ratio $X/Y$ (which is known to follow a doubly noncentral $F$ distribution, given the assumptions stated above).

The $r$-th raw moment $\mu'_r$ of $X/Y$ is defined as $E\left[\left(\frac{X}{Y}\right)^r\right]$. Using the independence of $X$ and $Y$, the first equality in equation (2.1) of the paper says (as far as I understand):

$$\mu'_r = E\left[X^r\right] E\left[Y^{-r}\right] \, \mathrm{,}$$

which makes sense to me.

My first doubt is a notation issue. The previous equation is actually typed in the paper as follows:

$$\mathtt{\mu'_r=EX^r \cdot EY^{-r}}$$

(using a typewriter font, as all the paper).

Later in the same page, the authors include this notation explanation:

$\mathtt{\mu_Y=EY}$

So I guess that $\mathtt{EY^{-r}}$ stands for $E\left[Y^{-r}\right]$, doesn't it?

The previous equation is further developed like this in the paper:

$$\mathtt{\mu'_r=EX^r \cdot EY^{-r} = EX^r \cdot E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r} } \, \mathrm{.}$$

And, maybe it is very easy, but it is here where I get totally lost. I do not understand this last step. Could you please give me some light?

NOTE: According to this thread: Is it okay to ask question about a specific paper / model?, it is OK to post questions on specific papers like the present one.

UPDATE

Just (hoping) to make it clearer, the whole equation (2.1) in the article is like follows (using their notation):

\begin{align*} \mathtt{\mu'_r} &\mathtt{= EX^r \cdot EY^{-r}} \\ &\mathtt{= EX^r \cdot E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}}\\ &\mathtt{= EX^r \cdot \left[ 1 + \binom{-r}{2}\frac{\mu_{2,Y}}{\mu_Y^2} + \binom{-r}{3}\frac{\mu_{3,Y}}{\mu_Y^3} + \binom{-r}{4}\frac{\mu_{4,Y}}{\mu_Y^4} + \cdots \right]} \end{align*}

where $\binom{-r}{k}$ stands for $\,\prod_{j=1}^{k}{\frac{-r-j+1}{j}}\,$, $\,\mu_Y = E[Y]\,$ and $\,\mu_{r,Y} = E\left[\left( Y - \mu_Y \right)^r\right]\,$ (central moments).

• It's not clear to me what's going on in the last equation. What do the authors do with it after they've obtained it? Maybe that gives some clues... – S. Catterall Sep 11 '17 at 17:12
• It seems too good (simple) to be true. Do computations give the same result? – Martijn Weterings Sep 11 '17 at 20:18
• @MartijnWeterings No, they don't, as far as I've checked. – Vicent Sep 12 '17 at 6:58
• I've added the whole equation to my question, just to try to make it clearer. – Vicent Sep 12 '17 at 7:28

## 1 Answer

Yes, $EY^{−r}$ stands for $E[Y^{−r}]$. (I dislike not making it explicit because it leaves too many opportunities for misunderstandings and errors.)

With respect to the later part, consider:

$Y=\frac{Y}{E(Y)}\cdot E(Y)= E(Y)\cdot [\frac{Y}{E(Y)}-1+1]= E(Y)\cdot [\frac{Y-E(Y)}{E(Y)}+1]$

Therefore

$EY^{-r} = [E(Y)]^{-r}\cdot E \left[ \left( 1 + \frac{Y-EY}{EY} \right)^{-r}\right] \, .$

So now compare with the paper. The equation in the paper is

$\mu'_r=EX^r \cdot EY^{-r} = EX^r \cdot E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}\,.$

Dividing through by $EX^r$ we have

$\mu'_r/EX^r = EY^{-r} = E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}$.

So we see from that last equality that they're asserting

$EY^{-r} = E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}\quad{^\ddagger}$

and so we can see $[E(Y)]^{-r}$ seems to have disappeared -- there's a term missing in the paper.

$^\ddagger$ -- keeping in mind that when they write $E\text{<term>}^{-r}$ it seems they intend $E(\text{<term>}^{-r})$, so this means $E\left( \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}\right)$

• The answer gives a possible origin for the equation. But is it a correct reasoning? Can we reason the following: if $Y = E(a) \cdot b$ then $E[Y^r] = [E(a)]^r \cdot E[b]^r$ or $E[Y^r] = [E(a)]^r \cdot E[b^r]$ ... ? Is cov(a,b)=0? I imagine this algebra with estimates of moments (especially negative) as much more complex. But maybe the $\chi^2$ makes it easier? That should be specified. To me the step $Y = E(Y) \cdot [\frac{Y-E(Y)}{E(Y)}+1]$ is trivial, but the part after 'therefore' is not. – Martijn Weterings Sep 12 '17 at 2:04
• I see now. The first line uses correctly $E(E(a) \cdot b) = E(a) \cdot E(b)$. But more is wrong (with the power terms). It is not just the $[E(Y)]^r$ term that disappeared. also the power in the $E[(1+\frac{Y-EY}{EY})^r]$ changed place. – Martijn Weterings Sep 12 '17 at 2:16
• No, I don't think the power changed place there either in the original exposition nor in mine, I believe their intent was $E([...]^{-r})$ not $(E[...])^{-r}$; it's just that the way the expected value was written was ambiguous; I deliberately removed the ambiguity at the end when I did it – Glen_b Sep 12 '17 at 2:32
• The equation in the paper is $\mu'_r=EX^r \cdot EY^{-r} = EX^r \cdot E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}$. Dividing through by $EX^r$ we have $\mu'_r/EX^r = EY^{-r} = E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}$. Dealing with that last equality removes a distracting term in $X$ and lets us work with a simple equation in $Y$, making it easier to spot the problem (i.e. to see that they have missed a term). – Glen_b Sep 12 '17 at 7:33
• Yes, that's what I think they should have. I didn't carefully examine the paper beyond checking that you correctly transcribed the equation in question (an important thing to double check), but my recollection is that the term didn't look like it was in the next line with the series expansion either. I also can't find any indication that there were any errata published later so it seems that the issue in those lines went unrecognized. It's quite possible that the error is fixed up at some later point in the paper though. – Glen_b Sep 12 '17 at 8:04