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Can you help me find the following:

$$E [W_1 \mid X(3) = 1]$$

$W_1$ is the time that the first event occurs, and $X(3) = 1$ signifies that one event occured before time 3. The rate of this question is lambda.

Here is what I have done so far:

I know that $$P(W_1 \le s \mid X(3) = 1) = P(X(s) \ge 1 \mid X(3) = 1) = 1 - P(X(s) = 0 \mid X(3) = 1) = 1-s/3$$

Differentiating this gives me the density $-1/3$.

Taking the integral from s to 3 of $-1/3$ gives me $1 - s/3$. Is this correct? I believe the answer should not be in terms of s so this leads me to believe I am wrong. Help would be great. Thanks a bunch.

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2 Answers 2

Until the point where you write $P(X(s)=0|X(3)=1)$, your proof is allright.

From there, you could write $P(X(s)=0|X(3)=1)$ as $A/B$ with $B=P(X(3)=1)$ and $$ A=P(X(s)=0,X(3)=1)=P(X(s)=0,X(3)-X(s)=1]. $$ By the independence properties of Poisson processes, the events $[X(s)=0]$ and $[X(3)-X(s)=1]$ are independent, hence $A=P(X(s)=0)P(X(3)-X(s)=1)$. Finally, $X(3)-X(s)$ is Poisson with parameter $(3-s)$, hence $$P(W_1<s|X(3)=1)=1-A/B=1-p_s(0)p_{3-s}(1)/p_3(1), $$ where, for every positive $t$ and positive integer $k$, $p_t(k)$ is the probability that a Poisson random variable with parameter $t$ is $k$. Thus, $p_t(0)=\mathrm{e}^{-t}$ and $p_t(1)=t\mathrm{e}^{-t}$, and $$ P(W_1<s|X(3)=1)=1-(3-s)/3=s/3. $$ This means that $W_1$ conditionally on $[X(3)=1]$ is uniform on $[0,3]$, hence $$ E(W_1|X(3)=1)=3/2. $$ More generally, conditionally on $[X(t)=k]$, the $k$ points of the Poisson process until time $t$ are uniformly distributed on the hypercube $[0,t]^k$. For this and much more, you could try the book Poisson Processes by J. F. C. Kingman.

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You clearly have a problem. Probability densities should be non-negative, and a cumulative distribution function should be (at least weakly) increasing.

In fact in a Poisson process, given $X(a)=n-1$ and $X(b)=n$, $W_n$ is uniformly distributed on $(a,b]$, therefore with density $\frac{1}{b-a}$ in this interval. The expected value of $W_n$ will be halfway between $a$ and $b$.

So going back to your problem, you should get $Pr(W_1 \le s|X_3=1) = \frac{s}{3}$ and its derivative is $\frac{s}{3}$. This will give you $E[W_1|X_3=1]=\frac{3}{2}$

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