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Suppose there are i.i.d. exponential random variables $x_1,\dots,x_n, y_1,\dots,y_m$. Can we write the following expectation $$E\left(\cfrac{a_1x_1+\dots+a_nx_n}{b_1y_1+\dots+b_my_m}\right),$$ as the sum of some functions of the coefficients i.e., of the form $$\cfrac{f_1(a_1)+\dots+f_n(a_n)}{g_1(b_1)+\dots+g_m(b_m)},$$ where $f_i$ and $g_j$ are functions and $a_i$ and $b_j$ are constants. I think the numerator can be written as the sum of expectations $a_1Ex_1+\dots+a_nEx_n$ and of course $Ex_i$ are all equal by i.i.d. I am not sure how to simplify the denominator.

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    $\begingroup$ Are the families $(x_i)_{1\leqslant i\leqslant n}$ and $(y_i)_{1\leqslant i\leqslant n}$ independent? $\endgroup$ – Davide Giraudo Dec 28 '13 at 17:22
  • $\begingroup$ @DavideGiraudo Yes. All random variables are mutually independent. I edited the question to make it clear. Thanks $\endgroup$ – triomphe Dec 28 '13 at 17:24
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    $\begingroup$ In this case, you use $\mathbb E[U/V]=\mathbb E[U]\cdot \mathbb E[1/V]$. The expectation of $U$, the numerator shouldn't be problematic. However, the computation of the expectation of the denominator seems more tricky. $\endgroup$ – Davide Giraudo Dec 28 '13 at 17:26
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    $\begingroup$ It's fairly routine to show that no generality is lost by taking $Ex_1=1$. In that case if $b_1=\cdots=b_m$, then $y_1+\cdots+y_m$ has a gamma distribution $y^{m-1} e^{-y}\,dy/\Gamma(m)$, and integrating $1/y$ times that is routine. If the $b$s are not all equal, you've got a harder problem. $\endgroup$ – Michael Hardy Dec 28 '13 at 17:29
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    $\begingroup$ Are you asking for an expression of the expectation as $(u(a_1)+\cdots+u(a_n))/(v(b_1)+\cdots+v(b_m))$, for some functions $u$ and $v$ independent of the parameters $a_i$ and $b_j$? $\endgroup$ – Did Dec 28 '13 at 18:59
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Assume that $m\geqslant2$, that every $b_k$ is positive, and, without loss of generality that every $y_k$ is standard exponential. Apply to the denominator the identity, valid for every $z\gt0$, $$ z^{-1}=\int_0^\infty\mathrm e^{-zs}\mathrm ds. $$ For $z=\sum\limits_kb_ky_k$, this yields $$ E[z^{-1}]=\int_0^\infty \prod_kE[\mathrm e^{-b_ky_ks}]\mathrm ds. $$ For every standard exponential $y$ and every positive $s$, $E[\mathrm e^{-sy}]=1/(1+s)$, hence $$ E[z^{-1}]=\int_0^\infty Q(s)\mathrm ds,\qquad Q(s)=\prod_k\frac1{1+b_ks}. $$ Assuming the real numbers $b_k$ are distinct, the decomposition of $Q$ reads $$ Q(s)=\sum_kc_k\frac{b_k}{1+b_ks}, $$ where one can note that $\sum\limits_kc_k=0$ since $sQ(s)\to0$ when $s\to\infty$. Thus, $$ E[z^{-1}]=\sum_k\left.c_k\log(1+b_ks)\right|_0^\infty=\sum_kc_k\log b_k. $$ To be complete, note that $$ c_k=b_k^{m-2}\prod_{j\ne k}\frac1{b_k-b_j}, $$ hence $$ E[z^{-1}]=\sum_k\left.c_k\log(1+b_ks)\right|_0^\infty=\sum_kb_k^{m-2}\log b_k\prod_{j\ne k}\frac1{b_k-b_j}. $$ This rules out the possibility of a formula such as the one you have in mind.

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  • $\begingroup$ Thank you. Would it be possible to lower bound this expression by one like the one I had in question? I mean $E[z^{-1}]>g_1(b_1)+\dots+g_m(b_m)$? $\endgroup$ – triomphe Dec 29 '13 at 0:02
  • $\begingroup$ Dunno. But note that in your question, you asked about $1/(v(b_1)+\cdots+v(b_m))$, not $v(b_1)+\cdots+v(b_m)$. $\endgroup$ – Did Dec 29 '13 at 0:03
  • $\begingroup$ Yes the mistake is in the above comment of mine. It should be $E[z^{-1}]>\frac{1}{g_1(b_1)+\dots +g_m(b_m)}$. Thanks $\endgroup$ – triomphe Dec 29 '13 at 0:11
  • $\begingroup$ the integral identity is nice! $\endgroup$ – Sandeep Silwal Jul 8 at 15:48

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