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I am trying to take the compound distribution of a uniform distribution on the interval $(0, \frac{t}{2})$ where $t$ is parameterized by the exponential distribution $\varepsilon (\lambda)$.

Taking the compound distribution I believe gives the following integral

$$\int_0^\infty \frac 2 t \lambda e^{-\lambda t} \, dt$$

I am not able to evaluate this integral. Have I made a mistake in my reasoning anywhere. Any help would be appreciated.

Michael Hardy
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Philip Lassen
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    Your question looks as if it might mean the following. You have a random variable $T$ whose distribution is $e^{-\lambda t}(\lambda,dt)$ for $t\ge0$ (and note how I distinguish between the meanings of capital $T$ and lower-case $t$) and another random variable --- let us call it $X$ --- whose distribution is uniform between $0$ and $T/2.$ You want to "take the compound distribution", which I'm guessing may mean you want to find the marginal distribution of $X.$ But the integral you've written is just a number; it does not identify any distribution. $\qquad$ – Michael Hardy May 22 '18 at 17:31
  • $($Actually the value of the integral $\displaystyle\int_0^\infty \frac 2 t \lambda e^{-\lambda t} , dt$ is $+\infty.) \qquad$ – Michael Hardy May 22 '18 at 17:32
  • Is my first comment what your question is intended to mean? – Michael Hardy May 22 '18 at 17:33
  • Yes your first comment is what I was struggling to say. Thanks for the clarification on the proper notation and terminology. – Philip Lassen May 23 '18 at 04:37

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First of all, notice that: $$P(X<x \mid T = t) = \begin{cases} \displaystyle\int_0^x\frac{2}{t}ds & \text{if } x<\displaystyle\frac{t}{2}\\ 0 & \text{if } x>\displaystyle\frac{t}{2}\\ \end{cases} = \begin{cases} \displaystyle\frac{2x}{t} & \text{if } x<\displaystyle\frac{t}{2}\\ 0 & \text{if } x>\displaystyle\frac{t}{2}\\ \end{cases}$$

Then, using the law of total probability, we get that:

$$P(X<x) = \int_0^{+\infty}P(X<x \mid T = t) f_T(t) \, dt,$$

where $f_T(t) = \lambda e^{-\lambda t}$ is the pdf of $T$. The previous integral is rewritten as follows:

$$P(X<x) = \int_{2x}^{+\infty} \frac{2x}{t} \lambda e^{-\lambda t} \, dt.$$

Unfortunately, the last integral can't be solved using known functions. Moreover, as pointed out by Michael Hardy, I'm not sure that it converges.

the_candyman
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