$X_1$ and $X_2$ are two independent random variables whose expected values are known. I am trying to find the expected value of $(X_1/(X_1+X_2))$. Since $X_1$ and $X_1+X_2$ are dependent, I tried to use the formula for covariance to calculate the expected value of the product but this results in variance ($X_1$) being part of the resultant expression which is unknown. Is there any other way of finding the expected value of the expression $(X_1/(X_1+X_2))$?
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3$\begingroup$ there's no general closed form formula $\endgroup$– AksakalMay 30, 2018 at 23:22
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1$\begingroup$ Can it be approximated? $\endgroup$– gagansoMay 31, 2018 at 2:43
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1$\begingroup$ We need some more information. Are $X_1, X_2$ positive random variables? In that case the ratio are $\le 1$. If not, the ratio could even have infinite expectation (or undefined). See stats.stackexchange.com/questions/299722/… $\endgroup$– kjetil b halvorsen ♦Jul 6, 2018 at 23:01
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$\begingroup$ Yes, they are positive random variables. $\endgroup$– gagansoJul 7, 2018 at 4:50
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1$\begingroup$ Your question simplifies by defining $Y=X_2/X_1,$ for it then asks for the expectation of $1/(1+Y)$ given only partial information about the expectation of $Y$ itself and no information about how it varies (apart from being a positive random variable). All that can be said at this level of generality is the trivial conclusion: the answer lies between $0$ and $1.$ $\endgroup$– whuber ♦Jan 24, 2020 at 14:35
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1 Answer
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An extended comment:
Let $\mathcal G(a,b)$ be the gamma density with pdf $f(x)\propto e^{-ax}x^{b-1}\mathbf1_{x>0}$.
Consider independent random variables $X_1\sim\mathcal G(a,b)$ and $X_2\sim\mathcal G(a,c)$.
Then it can be shown that $X_1+X_2$ is independent of $\frac{X_1}{X_1+X_2}$.
In fact, $X_1+X_2\sim\mathcal G(a,b+c)$ and $\frac{X_1}{X_1+X_2}\sim\mathcal{Be}(b,c)$, the beta distribution of the first kind.
This is a standard relation between beta and gamma variables.
Now,
\begin{align}E(X_1)&=E\left(\frac{X_1}{X_1+X_2}\cdot X_1+X_2\right) \\&=E\left(\frac{X_1}{X_1+X_2}\right)E(X_1+X_2) \end{align}
Hence, $$E\left(\frac{X_1}{X_1+X_2}\right)=\frac{E(X_1)}{E(X_1)+E(X_2)}$$
While I do not know a general formula for the expectation you ask for, in some special cases like the above, we can use well-known relations between functions of the random variables under consideration and use their independence and proceed. Here, of course all the expectations exist which may not be the case in general.
answered Jul 7, 2018 at 10:39