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Could someone give me some tips?

Let $e_1,e_2,\dots$ be iid normal mean 0 variance 1. Let $X_t := e_1+\cdots+e_t$, for $t=1,2,\dots$ and $X_0 := 0$. (So we have a discrete-time random walk whose steps are iid $N(0,1)$)

Define first-hitting time $T_0 := \inf\{t>0 : X_t <0\}$.

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What is the CDF of $T_0$, i.e. what is $\mathbb{P}(T_0 \leq t)$?

And second (probably more difficult) question: what is $\mathbb{E}( X_t \mid T_0 > t )$?

Thank you.

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Here is an answer for your first question.

This is easier than dealing with simple symmetric random walk.

When the steps are diffuse,

you don’t have to worry about *ties* or *parity*.

You are looking at a Brownian motion, evaluated at integer times $t$.

Then defining $$T_0=\inf(t > 0, t\in {\mathbb Z}: B_t < 0)$$ and integer $t > 0$

we have $$ P(T_0 > t)=P(B_1 > 0,B_2 > 0,\dots, B_t > 0)={2t\choose t}/4^t.$$

It turns out that the answer is the same for any random walk with

symmetric, diffuse increments.

This follows from Theorem 9.11 of Kallenberg’s *Foundations of Modern Probability (2nd edition)*, for example.

(A great book, by the way!)

The key insight is that you should not to try to solve the

probability exactly, but rather set up a recursion that the probabilities solve.

Setting $p(t)=P(T_0>t)$ for integer $t\geq 0$, you can show that

for any $t$ we have $\sum_{j=0}^t p(t-j)p(j)=1$.

The probabilistic justification for the recursion formula is that

every path of a diffuse, symmetric, discrete-time random walk can be broken into two

*positive* paths at the time $j$ where it achieves its minimum value.

You may also find Konrad Jacob’s book *Discrete Stochastics* to be useful.

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