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Let $a, b, c, d, e, f$ be complex numbers with nonnegative real parts and nonnegative imaginary parts, and let $X_{1}, X_{2}, X_{3}, X_{4}$ be independent standard normal random variables. How can I verify the following:

$$E\left[\frac{a X_{1} + b X_{2} + c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] = E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right]?$$

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    $\begingroup$ The expectation does not exist and so you cannot verify the expression you want. Note that even the ratio of two independent standard normal random variables is a Cauchy random variable for which the expectation is undefined and does not exist (no, not even in the sense that $f(x) = 1/|x|$ is undefined at $0$). See, for example, this question and its answers if you are unfamiliar with this notion. $\endgroup$ – Dilip Sarwate Jun 18 '15 at 13:28
  • $\begingroup$ The expression was used in a published article. The authors "exploited the independence of the $X_{i}$'s" to derive it. However, I can't seem to verify it. $\endgroup$ – thor Jun 18 '15 at 13:35
  • $\begingroup$ Is your question "How did they get rid of $cX_3+dX_4$ in the numerator?" I have no idea how the authors might have "exploited" the independence of the $X_i$'s to arrive at the alleged equality, and suspect that it is a proof by hand-waving that the reviewers and the editor accepted without thinking about the matter. I expect that Moderators whuber and/or Glen_b will soon post the definitive answer. $\endgroup$ – Dilip Sarwate Jun 18 '15 at 13:48
  • $\begingroup$ Yes, I wondered how did they get rid of $cX_{3} + dX_{4}$ in the numerator. I copied the expression verbatim from their paper. $\endgroup$ – thor Jun 18 '15 at 13:53
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    $\begingroup$ It would be helpful if you included in the question the link to the published paper. $\endgroup$ – Alecos Papadopoulos Jun 18 '15 at 15:19
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Here is a hand-waving "proof" of the alleged result:

\begin{align}E\left[\frac{a X_{1} + b X_{2} + c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] &= E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right] + E \left[\frac{c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right]\\ &= E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right]+0\\ &= E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right] \end{align} where the first equality follows from the linearity of expectation while the assertion that $$E \left[\frac{c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right]=0$$ uses the fact that $cX_3+dX_4$ and $eX_1+fX_2$ are independent zero-mean normal random variables, and so their ratio has a symmetric distribution (no dissension from me thus far) and thus has expected value $0$ (which I do not agree with but which will likely pass muster with most people and even some readers of stats.SE).

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  • $\begingroup$ I think the logic of this answer may suffer from the same logical defect as Alecos' answer: you still need to evaluate the expectation of $(aX_1+bX_2)/(eX_1+fX_2)$ before you can draw a valid conclusion. $\endgroup$ – whuber Jun 18 '15 at 17:05
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    $\begingroup$ @whuber Oh, I absolutely agree that $E[(aX_1+bX_2)/(eX_1+fX_2)]$ needs to be evaluated in order to conclude the"equality" to be proved, and I did not intend my sort of tongue-in-cheek answer to be taken seriously as a proof of the result. I was merely wanting to answer the OP's implicit question "I wondered how did they get rid of $cX_3+dX_4$" in the comments on the main question, and supply a reasoning for that disappearance that many might accept as valid. $\endgroup$ – Dilip Sarwate Jun 18 '15 at 18:00
  • $\begingroup$ I see--thank you for the clearing that up (+1). $\endgroup$ – whuber Jun 18 '15 at 19:03
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Assume that the variables are zero-mean, and jointly independent, without specifying a distribution. By linearity of the expected value,

$$E\left[\frac{a X_{1} + b X_{2} + c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] = E\left[\frac{a X_1 + b X_2}{e X_1 + f X_2}\right] + E\left[\frac{c X_3 + d X_4}{e X_1 + f X_2}\right]$$

Let's concentrate on the second quotient. Due to independence we can distribute the expected value in the second quotient, while using linearity again:

$$E\left[\frac{c X_3 + d X_4}{e X_1 + f X_2}\right]= \Big(c E(X_3) + d E(X_4)\Big)\cdot E\left( \frac {1}{eX_1 + fX_2}\right)$$

$$= \Big(c \cdot 0 + d \cdot 0 \Big)\cdot E\left( \frac {1}{eX_1 + fX_2}\right) = 0 \cdot E\left( \frac {1}{eX_1 + fX_2}\right)$$

since the variables are zero-mean...The fact that the coefficients are imaginary, changes nothing as regards multiplication by zero.

But what happens with the second expected value?

If indeed $X_1$ and $X_2$ are standard normal, then their sum is also normal, since they are independent. The reciprocal of a normal random variable does not have a first or a higher moment, they do not exist. So the product is not zero, and the assertion in the paper is not correct.

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  • $\begingroup$ Yes, it's enough for the coefficients to be positive constants. I can only access the article from my campus, since it has a paid subscription. $\endgroup$ – thor Jun 18 '15 at 16:01
  • $\begingroup$ This argument is not quite valid. It is possible to split any sum (which has a defined finite expectation) into two parts, each of which has no expectation. Thus, demonstrating that the second part of the sum has no expectation does not imply the original expression has no expectation. For example, let $X, Y, Z$ be independent, with $X$ and $Y$ of finite expectation but $Z$ with undefined expectation. Then $$X+Y=(X-Z)+(Y+Z)$$ is such a decomposition. $\endgroup$ – whuber Jun 18 '15 at 17:01
  • $\begingroup$ @whuber I am not sure that the "artificial" inclusion of terms that cancel out at all times is representative of the situations we are discussing here, where there exists a non-cancellable "problematic" component. And in case, how do we reconcile your comment with the linearity of the expected value? This linearity property is the linearity property of integrals, and as far as I can remember, it is valid to apply the decomposition, also on integrals that diverge or are undefined. Any insights would be appreciated. $\endgroup$ – Alecos Papadopoulos Jun 24 '15 at 9:39
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    $\begingroup$ The expectation is linear when it is defined. Until you have independently demonstrated that all the expectations you use are defined, you could (in principle) be performing invalid operations with zeros and undefined values. Thus, no result of your operations is logically sufficient to demonstrate anything about the original expectation until the well-definedness of these quantities is established. But this is exactly what you are trying to prove! The logic is thereby circular. $\endgroup$ – whuber Jun 24 '15 at 13:07

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