TITLE: Distribution of maximum of partial sums of independent random variables
QUESTION [7 upvotes]: I have been looking all over the net to find a way to work out a probability distribution of a maximum of partial sums of independent random variables, but to no avail. So I have decided to try to work it out for myself. Here are the results of this endeavour and I would appreciate if people with better understanding of probability than me would give it a look over to see if I’ve made a mess of it somewhere. Many thanks.  Here it goes.
Given a set $\{X_i:i=0,1,\ldots,n \}$ of independent random variables with $X_0 = 0$ and with given p.d.f.'s $f_{X_i}(x)$ and corresponding c.d.f's $F_{X_i}(x)$, define $S_k=\Sigma_{i=0}^k\,X_i$, and $M_k=\max\{S_0,S_1,\ldots,S_k \}$, and we want to find a distrinution of $M_n$. We have
$$
\begin{eqnarray*}
P(M_n<m)&=&P(\max\{M_{n-1},S_n\}<m)\\
&=&P(M_{n-1}<m,S_n<m)\\
&=&P(S_n<m|M_{n-1}<m)P(M_{n-1}<m),\\
\end{eqnarray*}
$$
where
$$
\begin{eqnarray*}
P(S_n<m|M_{n-1}<m)
&=&P(S_n<m|\max\{M_{n-2},S_{n-1}\}<m)\\
&=&P(S_n<m|S_{n-1}<m)\\
&=&P(S_n<m,S_{n-1}<m)/P(S_{n-1}<m),
\end{eqnarray*}
$$
where
$$
\begin{eqnarray*}
P(S_n<m,S_{n-1}<m)
&=&P(S_{n-1}+X_n<m,S_{n-1}<m)\\
&=&\int\limits_{-\infty}^m f_{S_{n-1}}(s)\int\limits_{-\infty}^{m-s}f_{X_n}(x)\mathrm dx\,\mathrm ds\\
&=&\int\limits_{-\infty}^m f_{S_{n-1}}(s)F_{X_n}(m-s)\mathrm ds,
\end{eqnarray*}
$$
where
$$
\begin{eqnarray*}
f_{S_k}(s)&=&\int\limits_{-\infty}^{\infty} f_{X_k}(x)\,f_{S_{k-1}}(s-x)\mathrm dx,
\end{eqnarray*}
$$
with
$$
\begin{eqnarray*}
f_{S_0}(s)=\delta (s),
\end{eqnarray*}
$$
so that putting it all together gives a recursion formula
$$
\begin{eqnarray*}
F_{M_n}(m) = \frac{F_{M_{n-1}}(m)}{F_{S_{n-1}}(m)} \int\limits_{-\infty}^m f_{S_{n-1}}(s)F_{X_n}(m-s)\mathrm ds
\end{eqnarray*}
$$
with
$$
\begin{eqnarray*}
F_{M_0}(m) = H(m)
\end{eqnarray*}
$$
the Heaviside step function.
Added 1:
Using another approach, I have obtained the following result
$$
f_{M_n}(m) = f_{M_{n-1}}(m)F_{X_n}(0) + \int\limits_0^m f_{X_n}(m-x)f_{M_{n-1}}(x)\mathrm dx
$$
which for 2 normal r.v. seems to give a result that agrees with the one put forward in one of the answers by Sasha.
Added 2:
Finally I got some free time to look at this problem again and here are my thought on it.  Once again, I would appreciate any comments on it.
We begin by considering a joint distribution of $f_{S_1,S_2}(s_1,s_2)$ where $S_1 = X_1$,  $S_2 = S_1 + X_2$, $X_1 \sim X_2 \sim X$, and $X_1$ and $X_2$ are independent. We have
$$
f_{S_1,S_2}(s_1,s_2) = f_{S_2 \mid S_1}(s_2 \mid s_1)f_{S_1}(s_1) = f_{X_2}(s_2 - s_1)f_{X_1}(s_1) = f_{X}(s_2 - s_1)f_{X}(s_1)
$$
Next, we have
$
\begin{eqnarray*}
F_{M_n}(m)&=&P(M_n<m)\\
&=&P(\max\{S_{n},S_{n-1},...,S_1\}<m)\\
&=&P(S_{n}<m,S_{n-1}<m,...,S_1<m)\\

&=&\int\limits_{-\infty}^m ... \int\limits_{-\infty}^m f_{S_n,S_{n-1},...,S_1}(s_n,s_{n-1},...,s_1)\mathrm ds_n \mathrm ds_{n-1} ... \mathrm ds_1 \\

&=&\int\limits_{-\infty}^m ... \int\limits_{-\infty}^m 
f_{S_n \mid S_{n-1},...,S_1}(s_n \mid s_{n-1},...,s_1)...
f_{S_2 \mid S_1}(s_2 \mid s_1)f_{S_1}(s_1)
\mathrm ds_n \mathrm ds_{n-1} ... \mathrm ds_1 \\

&=&\int\limits_{-\infty}^m ... \int\limits_{-\infty}^m 
f_{S_n \mid S_{n-1}}(s_n \mid s_{n-1})...
f_{S_2 \mid S_1}(s_2 \mid s_1)f_{S_1}(s_1)
\mathrm ds_n \mathrm ds_{n-1} ... \mathrm ds_1 \\

&=&\int\limits_{-\infty}^m ... \int\limits_{-\infty}^m 
f_X(s_n - s_{n-1})...
f_X(s_2 - s_1)f_X(s_1)
\mathrm ds_n \mathrm ds_{n-1} ... \mathrm ds_1 \\

&=& \prod_{i=1}^n \int\limits_{-\infty}^m f_X(s_i - s_{i-1}) \mathrm ds_i
\end{eqnarray*}
$$
where $s_0 \equiv 0$. I hope the above notation is clear enough. Now
$$
\begin{eqnarray*}
F_{M_n}(m) = \mathbb E[\mathbb I_{ M_n \leq m}] = \prod_{i=1}^n \int\limits_{-\infty}^{-\infty} f_X(s_i - s_{i-1}) \mathbb I_{s_i \leq m} \mathrm ds_i.
\end{eqnarray*}
$$
The characteristic function $\varphi_{M_n}(t) = \mathbb E[\mathbb e^{i t M_n}]$ is then
$$
\begin{eqnarray*}
\varphi_{M_n}(t) = \prod_{i=1}^n \int\limits_{-\infty}^{-\infty} f_X(s_i - s_{i-1}) \mathbb e^{i t s_i} \mathrm ds_i
\end{eqnarray*}
$$

REPLY [2 votes]: If you're interested in the situation for large $n$, then note that if your random variables $X_i$ have finite variance, then as $n \to \infty$ the scaling limit of $S_k$ is a Brownian motion $B_t$.  The distribution of $M_t = \max_{0 \le s \le t} B_s$ is known to be the same as that of $|B_t|$, i.e. the absolute value of a normal random variable with mean 0 and variance $t$, and the pdf is
$$P(M_t \in dx) = \sqrt{\frac{2}{\pi t}} e^{-x^2/2t} dx.$$
See, e.g., I. Karatzas and S. Shreve, Brownian Motion and Stochastic Calculus, 2ed, Proposition 2.8.1 (the result quoted above is Problem 2.8.2 but you just have to take the result in Proposition 2.8.1 and integrate out $a$).