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I am wondering if the following expectation can be computed in the shown manner

$$\begin{aligned} E[\log_2(1+X) \mid X>a] &= \int_{0}^{\infty} P( \log_2(1+X)>t \mid X>a) dt \\ &= \int_{0}^{\infty} P(X>2^t-1 \mid X>a) dt \\ &= \int_{0}^{\infty} \frac{P(X>2^t-1 \cap X>a)}{P(X>a)} dt \\ &= \frac{1}{P(X>a)} \left(\int_{0}^{a} P(X>a) dt + \int_a^{\infty} P(X>2^{t}-1) dt \right) \end{aligned}$$

Note that there is no closed form expression for the PDF of X hence I am trying to find alternative ways to compute the expectation.

Thank you

Lee David Chung Lin
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hgm
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  • If there is no closed form for the density of $X$, do you have a closed form for $\mathbb P(X>a)$? If not, I don't see how this helps. – Math1000 Sep 05 '19 at 22:10
  • I dont have a close form for $P(X>a)$ however, I can compute it numerically using matlab or other programming languages – hgm Sep 05 '19 at 22:12
  • I just want to know if the math is correct in terms of expectation computations – hgm Sep 05 '19 at 22:13
  • I don't think the first equality is valid. In general, $\mathbb E[f(X)\mid X>a] \ne \int_0^\infty \mathbb P(f(X)>t\mid X>a)\ \mathsf dt$. – Math1000 Sep 05 '19 at 22:23
  • Also I suppose you're assuming that $\mathbb P(X\geqslant 0)=1$, hence the lower bound on the integral of $0$? – Math1000 Sep 05 '19 at 22:24
  • yes the lower bound is $0$, any ideas how to approach this conditional expectation then? – hgm Sep 05 '19 at 22:30
  • @math1000, I am just wondering if you are sure that $E[f(c) |x>a] \neq \int_0^{\infty} P(f(x) >t|x>a) dt$? Can you help pinpoint useful references? – hgm Sep 06 '19 at 23:15
  • According to the last answer in this post https://math.stackexchange.com/questions/2693408/conditional-expectation-of-x-given-x-1, the method I detailed should work, can someone confirm? – hgm Sep 06 '19 at 23:52
  • Let $X$ be exponentially distributed with parameter $\lambda$, then the moment generating function is $$\mathbb E[e^{tX}] = \int_0^\infty e^{tx} \lambda e^{-\lambda x}\ \mathsf dx = \frac\lambda{\lambda-t}, t<\lambda$$ but $$ \int_0^\infty \mathbb P(e^{tX}>x)\ \mathsf dx = \int_0^\infty \mathbb P(X>t^{-1}\log x)\ \mathsf dx = \int_0^\infty x^{-\lambda/t}\ \mathsf dx$$ which does not converge. – Math1000 Sep 07 '19 at 06:15
  • Or let $X$ have uniform distribution on $(0,1)$, then $$ \mathbb E[e^{tX}] = \int_0^1 e^{tx}\ \mathsf dx = \frac1t(e^t-1), $$ but $$ \int_0^1 \mathbb P(X>e^{tX})\ \mathsf dx = \int_0^1 (1 - t^{-1}\log x)\ \mathsf dx = 1 + \frac1t. $$ – Math1000 Sep 07 '19 at 06:21

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