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

$$ \begin{eqnarray*} P(M_n<m)&=&P(\max\{M_{n-1},S_n\}<m)\\ &=&P(M_{n-1}<m,S_n<m)\\ &=&P(S_n<m|M_{n-1}<m)P(M_{n-1}<m),\\ \end{eqnarray*} $$


$$ \begin{eqnarray*} P(S_n<m|M_{n-1}<m) &=&P(S_n<m|\max\{M_{n-2},S_{n-1}\}<m)\\ &=&P(S_n<m|S_{n-1}<m)\\ &=&P(S_n<m,S_{n-1}<m)/P(S_{n-1}<m), \end{eqnarray*} $$


$$ \begin{eqnarray*} P(S_n<m,S_{n-1}<m) &=&P(S_{n-1}+X_n<m,S_{n-1}<m)\\ &=&\int\limits_{-\infty}^m f_{S_{n-1}}(s)\int\limits_{-\infty}^{m-s}f_{X_n}(x)\mathrm dx\,\mathrm ds\\ &=&\int\limits_{-\infty}^m f_{S_{n-1}}(s)F_{X_n}(m-s)\mathrm ds, \end{eqnarray*} $$


$$ \begin{eqnarray*} f_{S_k}(s)&=&\int\limits_{-\infty}^{\infty} f_{X_k}(x)\,f_{S_{k-1}}(s-x)\mathrm dx, \end{eqnarray*} $$


$$ \begin{eqnarray*} f_{S_0}(s)=\delta (s), \end{eqnarray*} $$

so that putting it all together gives a recursion formula

$$ \begin{eqnarray*} F_{M_n}(m) = \frac{F_{M_{n-1}}(m)}{F_{S_{n-1}}(m)} \int\limits_{-\infty}^m f_{S_{n-1}}(s)F_{X_n}(m-s)\mathrm ds \end{eqnarray*} $$


$$ \begin{eqnarray*} F_{M_0}(m) = H(m) \end{eqnarray*} $$

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

$$ \begin{eqnarray*} \varphi_{M_n}(t) = \prod_{i=1}^n \int\limits_{-\infty}^{-\infty} f_X(s_i - s_{i-1}) \mathbb e^{i t s_i} \mathrm ds_i \end{eqnarray*} $$

  • $\begingroup$ Here's a $\TeX$ tip---something you should learn if you're going to make a habit of using $\TeX$ or $\LaTeX$. When you write \max rather than \mathrm{max} or the like, the use of a standard operatorname not only prevents italicization (as \mathrm also does) but also causes standard spacing conventions to be followed (so you won't see $\text{max}A$ with no space) but furthermore causes all other standard conventions that go along with that operatorname to be followed, such as this: $\displaystyle\max_{i\in A} f(i)$. $\endgroup$ Sep 29, 2011 at 20:53
  • $\begingroup$ Test edit. I tried several times to prepend "@MichaelR" to the comment above. The software won't let me! And it didn't work in this test either. $\endgroup$ Sep 29, 2011 at 20:58
  • 1
    $\begingroup$ @Michael: I believe that someone mentioned this before. It seems that since the default recipient is the OP, it removes ${@}$OP. $\endgroup$
    – robjohn
    Sep 29, 2011 at 22:37
  • $\begingroup$ MichaelR, I've updated my answer - may be interesting for you to take a look. I'm not sure if updates of answers falls in inbox $\endgroup$
    – SBF
    Sep 30, 2011 at 8:49
  • $\begingroup$ The formula added after comments holds in one case and one case only: when $n=1$. $\endgroup$
    – Did
    Sep 30, 2011 at 18:29

4 Answers 4


The step where you assert that $$ \mathrm P(S_n<m\mid M_{n-1}<m)=\mathrm P(S_n<m\mid S_{n-1}<m) $$ because $(S_n)_{n\ge0}$ is a Markov chain, is a beautiful example of the subtle errors one can make when using the Markov property.

To see that this equality is false in general, consider an inhomogenous random walk $(S_n)$, exactly like in your post. Fix a time $n\ge2$, and assume that $X_{n-1}=-1$ or $-2$ but that $X_k=\pm1$ or $0$ for every $0\le k\le n$ such that $k\ne n-1$.

Then, $X_{n-1}\le0$ almost surely hence $[M_{n-1}<1]=[M_{n-2}<1]$. And $X_{n-1}\le-1$ almost surely hence $[M_{n-2}<1]\subseteq[S_{n-2}<1]\subseteq[S_{n-1}<0]$. And $X_n\le1$ almost surely hence, conditionally on $[S_{n-1}<0]$, $S_n<1$ almost surely. This shows that $$ \mathrm P[S_n<1\mid M_{n-1}<1]=1. $$ On the other hand, $[S_{n-1}=0]$ has positive probability because $n\ge2$, and $[S_{n}=1]$ contains $[S_{n-1}=0]\cap[X_{n}=1]$, hence $$ \mathrm P[S_n<1\mid S_{n-1}<1]<1. $$ Added in note The key properties of the random walk in this counterexample are that $X_{n-1}\le-x$ and $X_n\le x$ almost surely, for a given positive $x$, and that both events $[S_{n-1}=0]$ and $[X_n\ge1]$ have positive probability.

  • $\begingroup$ Yes, that's the bit that worried me the most and did feel to me like there was an error in it and that's why I've put it up here. I'll need to digest your answer a bit, but that will have to wait until tomorrow (as will the up-vote). Thank you very much for reading and replying to my post. I would also appreciate any references you might recommend on this topic. $\endgroup$
    – Confounded
    Sep 29, 2011 at 20:38
  • $\begingroup$ Probability with Martingales by David Williams. $\endgroup$
    – Did
    Sep 29, 2011 at 20:47
  • $\begingroup$ .@Didier Piau Given that $M_{n-2}=\max{\{M_{n-3},S_{n-2}\}}$ should your "$[M_{n-2}<1]\subseteq[S_{n-2}<1]$" not be $[S_{n-2}<1]\subseteq[M_{n-2}<1]$ instead? Thanks $\endgroup$
    – Confounded
    Sep 30, 2011 at 8:12
  • $\begingroup$ No: if $x=\max\{y,z\}$, then $x\ge y$ hence $x<1$ implies $y<1$. $\endgroup$
    – Did
    Sep 30, 2011 at 9:43
  • $\begingroup$ But $x < 1$ is a set $A = (-\infty, 1)$, while $y \leq x$ is a set $B = (-\infty, x)$, and $B \subseteq A$ or am I being extra thick? $\endgroup$
    – Confounded
    Sep 30, 2011 at 10:28

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 $a$).


So, there wrong move in your answer was already pointed out and the solution in terms of joint probabilities was discussed. Let us take a look at this problem from Markov's prospective. As I told you, the process $S_k$ is an inhomogeneous (time-dependent) Markov process - indeed, if you know the current state $S_k = s$ and the current time instance $k$, you know the distribution of $S_{k+1}$ even without knowledge of the whole history.

For the moment let us forget that the process is time-dependent, so we consider just a sum of iid random variables. Let us denote $u_n(m) = \mathsf P_0\{M_n<m\}$. Here $\mathsf P_0$ is a probability conditioned on the fact that you start from $S_0=0$. Due to the fact that $S$ is just a sum of independent increments, if we start not from $S_0=0$ but from $S_0 = s$ we will get $$ \mathsf P_s\{M_n<m\} = \mathsf P_0\{M_n<m-s\} = u_n(m-s).\quad (*) $$ That means that knowledge study $\mathsf P_s\{M_n<m\}$ for any $s$ and a fixed $m$ is equivalent to the knowledge of $\mathsf P_0\{M_n<m\}$ for any $m$, so that of the distribution of $M_n$. For the shortcut we denote $v_n(s):=u_n(-s)$.

It's more convenient to deal with $\mathsf P_s$ due to the following reason. Suppose, you know $v_n(s)$ - that is the probability that starting from $S_0 = s$ the maximum will not exceed the zero level, i.e. just put $m=0$ in $(*)$. How can you calculate now $v_{n+1}(s)$? So you start from $s$ and then jump to some point $t$ where you already know the probability to stay always inder the zero level - so you just need to apply the law of total probability: $$ v_{n+1}(s) = \int\limits_\mathbb R v_n(t)f_X(t-s)\,dt. $$ with $v_0(s) = 1-H(s)$. So you get from $(*)$ that $v_n(s) = \mathsf P_0\{M_n<-s\}$ for any $s\in \mathbb R$ and you know the distribution of the maximum.

What will change it you allow $X_i$ to have different distributions as in your original question? The additional variable will appear, but the idea of $(*)$ leaves unviolated: you still have the sum of independent increments. If you shift the starting point, you can shift the upper bound for the maximum on the same value and the probability will not change. However, what is tricky here is that the starting time matters. Suppose you know $$ v_n(0,s) = \mathsf P\{M_n<0|S_0=s\} $$ where $0$ in $v_n$ tells us that you conditioning on the value of $S_k$ at the moment $k=0$. But then to calculate $v_{n+1}(0,s)$ we need to know $v_n(1,s)$ since it will be different due to the differ in distributions of jumps. So, $$ v_{n+1}(0,s) = \int\limits_\mathbb R v_n(1,t)f_{X_1}(t-s)\,dt \quad (**) $$ with an initial condition $$ v_0(k,s) = 1-H(s) $$ for all $k = 0,1,2,3,...$

For example, to calculate $\mathsf P\{M_n<m|S_0=0\} = v_n(0,-m)$ you algorithm is:

  1. you know $v_0(n,s)=1-H(s)$ for any $s$.

  2. from $(**)$ you find $v_1(n-1,s)$;

  3. from the previous step you find $v_2(n-2,s)$ by applying $(**)$.


  4. from the previous step you find $v_{n-1}(1,s)$ and applying $(**)$ you've done.


Since $\max(S_0, \ldots, S_n) \le m$ implies $S_0 \le m \land S_1 \le m \land \ldots \land S_n \le m$, we have $$ \mathbb{P}(M_n \le m) = \mathbb{P}(S_0 \le m \land S_1 \le m \land \ldots \land S_n \le m) $$

That is the cdf of the maximum is the cdf of joint distribution at vector with equal components: $$ F_{M_n}(m) = F_{S_0,S_1,\ldots,S_n}\left(m,m,\ldots,m\right) $$ Your success in dealing with this will depend on tractability of $F_{S_0,\ldots, S_n}$. Assume $m \ge 0$, so that $0 = S_0 \le m$ is trivially true.

Considering $n=2$, and the case of standard normal $X_i$, vector $\{S_1, S_2\}$ follows binormal distribution with $\mu_1 = \mu_2 = 0$ and $\sigma_1 = 1$, $\sigma_2 = \sqrt{2}$ and correlation $\rho = \frac{1}{\sqrt{2}}$.

Use the fact that $\frac{d}{d x} F_{\mathcal{BN}}(x,y) = f\left(\frac{x-\mu_1}{\sigma_1}\right) \Phi\left( \frac{\sigma _1 \left(y-\mu _2\right) - \rho \, \sigma _2 \left(x-\mu _1\right)}{ \sigma _1 \sigma _2 \sqrt{1-\rho ^2}} \right) $, and similarly for derivative with respect to $y$, we have $$ \frac{\mathrm{d}}{\mathrm{d} m} \Phi_{S_1,S_2}\left(m,m\right) = \frac{1}{2} \left( f(m) + 2 f\left(\frac{m}{\sqrt{2}}\right) \Phi\left(\frac{m}{\sqrt{2}} \right)\right) $$ where $f$ is the pdf of standard normal, and $\Phi$ is its cumulative distribution function. This shows, that $\max(S_1, S_2)$ follows an equal proportion mixture between standard normal distribution and Azzalini's skew-normal distribution.


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