# Expected first hitting time for compound Poisson process?

Is there an analytic method to compute the expected first hitting time of an IID jump process with exponentially distributed time steps and jump sizes given by a Poisson random variable multiplied by a random variable that takes values {1,-1} with probability 0.5?

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The hitting time of what set? If it's a single particular integer then the answer is obvious (no calculation at all is needed) but unenlightening! – cardinal Aug 17 '12 at 20:22
I am interested in the hitting time of a single scalar value. Is it obvious? I must be missing something. :/ – user2303 Aug 17 '12 at 20:56
The process is nondecreasing, takes only integer values and can skip (jump over) any fixed integer. Conclusion? – cardinal Aug 17 '12 at 21:00
Ah, one thing I forgot: the Poisson jump size is also multiplied by a random variable X that equals 1 with probability 0.5 and -1 with probability 0.5. – user2303 Aug 17 '12 at 21:05
That changes matters a bit. Please edit your question to reflect your comment; I'll be happy to upvote it. – cardinal Aug 17 '12 at 21:18

The expected value is infinite.

For simplicity, replace the time between steps by a constant. This doesn't affect finiteness.

The expected number of steps before you cross the value $v$ is the sum of the probabilities that you haven't crossed $v$ by step $t$ for $t=0,1,2,...$.

The probability that the random walk has crossed $v$ by step $t$ is less than or equal to the probability that the random walk has had the same sign as $v$ at some time up to $t$.

For any set of IID random variables with distributions symmetric about $0$, the probability that the first $n$ partial sums are all positive is at least the value for continuous distributions, which is ${2n \choose n}/4^n$. See the answers to this MO question "A random walk with uniformly distributed steps."

By Stirling's approximation, ${2n \choose n}/4^n \sim c/\sqrt n$. The sum of $1/\sqrt{n}$ diverges, so the expected time before you land on the same side of the origin as $v$ is infinite, so the expected first crossing time is infinite.

This seems like it should be a standard result, so there may be a simpler proof, but I like the ${2n \choose n}/4^n$ result.

Edit: Here is a second proof using a version of the Optional Stopping Theorem. Again replace the waits between jumps by constants. The value is a martingale. Call it $(X_i)$. Let $T$ be the first time that $X_i$ hits or crosses the target $v$. If the conditions of the Optional Stopping Theorem hold,

1. $E[T] \lt \infty$

2. $\exists c ~\forall i ~E[|X_i - X_{i-1}| \bigg| X_0, X_1, ... X_{i-1}] \lt c$

then we can conclude $E[X_0] = E[X_T]$. This is not true, so either condition 1 or condition 2 fails. $E[|X_i - X_{i-1}| \bigg| X_0, ..., X_i] = E[|X_1 - X_0|] \lt \infty$ since magnitudes of the steps are IID Poisson variables, so condition 2 holds. This means condition 1 must not hold, and $E[T] = \infty$.

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(+1, hours ago) Your edit uses the argument I had in mind after the OP's update to the question. – cardinal Aug 18 '12 at 16:37