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Suppose we have a continuous-time stochastic process $X(t)$, which consists of a sequence of delta functions that, at each time $t$, have a probability $p(t)$ of taking a non-zero value. $p(t)$ lies in $[0,1]$ for all $t$ and just a scaled survival function of an exponential distribution, so $p(t)$ is decreasing in time. We are interested in the random variable $Y = \int_0^\infty X(t) f(t) dt$ where $f(t)$ is a (deterministic) decreasing function of time. Is it possible to take the expectation of this integral; that is, $E\left[Y\right] = E\left[\int_0^\infty X(t) f(t) dt\right]$?

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It is not immediately clear from your description that the integral would exist in any reasonable sense, much less that one would be able to take an expectation. Why should, for example, $\{t: X(t,\omega) > 0 \}$ be a Lebesgue measurable set for any fixed $\omega$? Perhaps I have missed some subtlety in your description. –  cardinal Jan 12 '13 at 22:13
    
Thank you for your help, @cardinal, and I don't think you have missed any subtlety in my description -- this problem goes a bit beyond my undergraduate training in statistics, so let me explain the problem more fully. I have a poisson process of rate $\lambda$ that lasts until $t=U$, where $U$ is a draw from an exponential distribution. Each event in the process is a purchase of (deterministic) value m. The random variable I am interested in is the discounted value of all purchases. That is, when a purchase at time $t$ is made, the value is $m \times exp(\delta t)$, to account for discounting –  Eli Stein Jan 15 '13 at 15:23
    
where $\delta$ is a discount factor. I want to take the expected value of this random variable, call it $S$. I am having a hard time finding the expected value of $S$. One way I tried to formulate this problem was defining $S$ as the integral above, where $X(t)$ takes the value $1$ if a purchase is made "epsilon close" to $t$ and $f(t) = exp(\delta t)$. Now, if the expectation operator could be passed through the integral (which, I know, it can't), this problem can be solved as $E[X(t)] = \lambda \times dt \times S(t)$, where $S(t)$ is the survival function for an exponential distribution. As –  Eli Stein Jan 15 '13 at 15:32
    
the expectation operator can't be passed through the integral, however, I am a bit stuck. Any ideas? Thank you for your help! –  Eli Stein Jan 15 '13 at 15:32
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