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I have a standard Poisson process $N(t)$ with arrival rate $\lambda$ per second. Each time an arrival occurs, a fair coin is flipped and a counter is increased or decreased by $1$ depending on the result. If we let $M(t)$ indicate the counter value at time $t$ and let $M(0) = 0$ (just like $N(0) = 0)$, at time $t$, if we knew that $N(t) = n$, then the value of the counter $M(t)$ could be understood as

$$ \# \text{(successes)} - \# \text{(failures)}$$

in $n$ Bernoulli trials with success probability $\frac{1}{2}$ . Using this logic, i need to find the conditional PMF $$\mathrm{Pr}[M(t) = m\mid N(t) = n]$$ and the expression for the PMF of $M(t)$.

I am having trouble on how to approach the problem and how # successes - # failures could be understood in finding the PMF. Any help would be greatly appreciated. Thanks!

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The probability for $k$ successes in $n$ Bernoulli trials with success probability $\frac12$ is $\binom nk2^{-n}$. The resulting value of $m$ is $k-(n-k)=2k-n$. Thus the conditional probability mass function is

$$ \def\pr{\operatorname{Pr}}\pr[M(t)=m\mid N(t)=n]=\begin{cases}\binom n{(n+m)/2}2^{-n}&n+m\text{ even,}\\\\0&\text{otherwise.}\end{cases} $$

Then by the law of total probability,

$$ \pr[M(t)=m]=\sum_{n=0}^\infty\pr[M(t)=m\mid N(t)=n]\pr[N(t)=n]\;, $$

and substituting $\pr[N(t)=n]=\mathrm e^{-\lambda t}(\lambda t)^n/n!$ yields an expression for $\pr[M(t)=m]$ as a sum over the values of $n$ with the same parity as $m$.

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    Great thanks! it makes much more sense now. What I'm having trouble finding now is the expected value of $M(t)$.2012-12-14
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    @Janson If your aim is to compute $E(M(t))$, (1) you should have said so in the question, and (2) none of the above is necessary.2012-12-15
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    @did so how would you do it then?2012-12-15
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    Since the probability of each event being a failure or a success is the same, `E(M(t)) = 0`.2012-12-16
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    @Janson Why the change of name?2012-12-16