# Limit of passage times of 1-dim random walk

## Homework Statement

$$T_{x}$$ is the passage time min{n : x(n) = x} for paths starting from x(0)=0

Show:
$$\lim_{x\rightarrow +\infty}E_{0}(e^\frac{-\alpha\\T_{x}}{x^2}) = e^\-\sqrt{2\alpha}$$

for any $$\alpha$$ ≥ 0

## The Attempt at a Solution

I came up with the recurrence relation of T_{x} as: $$T_{x} = T_{x-1}+T_{1} = xT_{1}$$

so $$E_{0}(e^\frac{-\alpha\\T_{x}}{x^2})$$ becomes $$E_{0}(e^\frac{-\alpha\\T_{1}}{x})$$

but that doesn't feel right

I also tried $$f(x) = \frac{e^\alpha}{2}(f(x+1)+f(x-1))$$ and $$\frac{T_{x}}{x^2} = \frac{\sum_{i=1}^{x}T_{1}^(i)}{x^2}$$ but again, don't know what to do from there

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Mark44
Mentor
What is E0?

E0 is the expectation of hitting zero