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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
up vote 2 down vote accepted

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

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