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With $X_1$, $X_2$ and $X_3$ being independent random variables, how can I compute $\mathbb{E}\left[ \frac{X_1}{X_1+X_2+X_3}\right]$?

Is $\mathbb{E}\left[ \frac{X_1}{X_1+X_2+X_3}\right] = \frac{\mathbb{E}\left[X_1\right]}{\mathbb{E}\left[X_1\right]+\mathbb{E}\left[X_2\right]+\mathbb{E}\left[X_3\right]}$? If not, how is it calculated?

Thank you in advance for any clarification.

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    $\begingroup$ Are you also assuming the variables all have the same distribution? $\endgroup$ – whuber Mar 24 at 18:02
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There is a famous "folk theorem" residual to my undergrad classes, namely that $$\mathbb E[X/Y] = \mathbb E[X] \big/ \mathbb E[Y]$$ and it is often permeates during exams, as it makes computing much easier. Sadly, the equality does not hold in general, even when $X$ and $Y$ are independent (due to Jensen's inequality).

In the case of $\mathbb E[X_1/X_1+X_2+X_3]$, numerator and denominator are dependent, which usually makes the computation more difficult. However, in the very special case when the three $X_i$'s are iid, $X_i/X_1+X_2+X_3$ has the same distribution for all three $i$'s and this leads to an obvious conclusion concerning the expectation of any of them. Assuming this expectation exists, of course. A counterexample is provided by a triplet of Normal variables (see Marsaglia's paper in connection).

As a special case where the identity works, take the Dirichlet $\mathcal D(\alpha_1,\ldots,\alpha_d)$ distribution, whose expectation is $$\mathbb E[Y_i]=\alpha_i\Big/\sum_{j=1}^d \alpha_j$$ One representation of a Dirichlet random vector $(Y_1,\ldots,Y_d)$ is $$Y_i=\frac{X_i}{X_1+\ldots+X_d}\qquad X_i\sim\mathcal G(\alpha_i,1)$$ where the $X_i$'s are independent. In that case, $$\mathbb E[Y_i]=\mathbb E[X_i\big/X_1+\ldots+X_d]= \mathbb E[X_i]\big/\mathbb E[X_1+\ldots+X_d]$$

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    $\begingroup$ The identity will work whenever the $X_i$ are iid with nonzero expectation, as you basically pointed out in an earlier version of this answer. $\endgroup$ – whuber Mar 24 at 18:29
  • $\begingroup$ @whuber: Is this enough to ensure that $X_1/(X_1+X_2)$ has a well-defined expectation? $\endgroup$ – Xi'an Mar 24 at 19:51
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    $\begingroup$ Good question: I don't think so. For instance, let the distribution be uniform on the values $\{-1,1,3\}.$ $|X_1/(X_1+X_2)|$ equals one divided by zero with a chance of $2/9.$ Continuous approximations to this will have comparable problems. Notice that these random variables are (a) bounded and (b) have zero probability to be zero. $\endgroup$ – whuber Mar 24 at 20:55
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    $\begingroup$ The expression like $X_1/(X_1 + X_2)$ also occured in this question stats.stackexchange.com/a/399952 . There is an expression from Hinkley for the case that the $X_i$ are Gaussian distributed (and the expectation will be infinite). In this question stats.stackexchange.com/a/438402 an intuitive view is given for the ratio distribution (and you could do the same for the correlated case). You could also express the distribution of the angle. And the expectation of the ratio is the expectation of the tangens of the angle, which becomes infinite when 90 deg has non-zero density. $\endgroup$ – Sextus Empiricus Mar 25 at 8:01
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    $\begingroup$ When the $X_i$ are continuous and non-negative then the division by 0 occurs only in the point $(X_1,X_2) = (0,0)$ (which has zero probability) and for the other values of $X_1,X_2$ the value of the ratio $X_1/(X_1+X_2)$ is between 0 and 1, such that the ratio won't have infinite or undefined expectation. $\endgroup$ – Sextus Empiricus Mar 25 at 8:18

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