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$$X\ \sim Exp(U) ~ and\ U\ \sim U(0,1) $$

The question asked for the value of $ P(X\geqslant 1)$

I saw the solution and it went like this:

$$P(X\geqslant 1) = E[P(X\geqslant 1)|U] = E[e^{-u}] = \int_{0}^{1}e^{-x}dx $$

Could someone explain the solution to me?

More specifically how did the solution derive each step?

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1 Answer 1

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You are given that the conditional distribution of $X$ given $U=u$ is $\text{Exp}(u)$ (presumably $u$ is the rate parameter), where $U$ itself is uniformly distributed on $(0,1)$.

The solution is an application of the law of total expectation, or equivalently, the law of total probability.

Suppose $I(X\ge 1)$ is an indicator variable that equals $1$ if $X\ge 1$ and equals $0$ otherwise. And let the density of $U$ be $f_U$. Then,

\begin{align} P(X\ge 1)&=E\left[I(X\ge 1)\right] \\&=E\left[E\left[I(X\ge 1\mid U\right]\right] \\&=E\left[P(X\ge 1\mid U)\right] \\&=E\left[e^{-U}\right] \\&=\int e^{-x}f_U(x)\,dx \\&=\int_0^1 e^{-x}\,dx \end{align}

In the penultimate step, law of the unconscious statistician was used to find the expectation.

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  • $\begingroup$ Yes I think I understand this better now. Would it be equally accurate if I wrote it this way? P(X≥1) =∫P(X≥1|U)fu(u) du = E[P(X≥1∣U)] $\endgroup$
    – Grey Han
    Apr 24 at 2:34
  • $\begingroup$ You can directly say $P(X\ge 1)=\int P(X\ge 1\mid U=x)f_U(x)\,dx$ by law of total probability. $\endgroup$ Apr 24 at 5:45

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