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Can someone explain this solution to me? The question was find E[W1 | X(t) = 2] where W1 is the time until the first event occurs and X(t) is a Poisson process. V1 represents the smallest of the uniform variables that are a part of this theorem. I understand the first line but I am unsure why they proceed to P(V1>s) and how they get (t-s/t)^2. I am also confused with how they calculate the expectation. Thanks for the help!

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up vote 5 down vote accepted

They are calculating $\renewcommand{\Pr}{\mathbb{P}} \Pr(V_1 > s)$ because $V_1$ is nonnegative and has the same distribution as the minimum of two i.i.d. random variables, so this quantity turns out to be convenient in two ways.

First, $\Pr(V_1 > s)$ can be easily employed to calculate the expected value of $V_1$. For any nonnegative random variable $X$, $\mathbb{E}(X) = \int_0^\infty \Pr(X > x) \,\mathrm{d}x$.

Now, since $X(t)$ is a Poisson process and we are conditioning on the fact that $X(t) = 2$, then the points of arrival are distributed as the order statistics of two uniform random variables on $[0,t]$. Thus, $V_1 = \mathrm{min}(U_1, U_2)$, equality here being in distribution, where $U_1$ and $U_2$ have a density that is $1/t$ on $[0,t]$ and 0 on $\mathbb{R} \setminus [0,t]$.

Hence, $$ \Pr(V_1 > s) = \Pr(U_1 > s, U_2 > s) = \Pr(U_1 > s) \Pr(U_2 > s) = \big(\Pr(U_1 > s)\big)^2 \>. $$

Finally, $$ \Pr(U_1 > s) = \int_s^t \frac{1}{t} \mathrm{d}u = \frac{t - s}{t} \> . $$

From here the rest is easy.

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