Cumulative sum over random numbers gives interesting results

Question

Recently, I encountered an interesting thing with numbers. I couldn't find a satisfactory explanation hence, I decided to post it here.

So, basically say we generate 5 random numbers with mean = 0 and standard deviation = 1 and take a cumulative sum over it and take the index which gives us the max value. To explain this with an example :

5 random numbers with mean = 0 and sd = 1

1.8871962   -2.0479581   -0.7508212   -1.2745548    -0.7129499

Take cumulative sum over it which gives us

1.8871962   -0.1607619    -0.9115832     -2.1861380   -2.8990878

and now we take the index of the highest value which is : 1 in this case.

Now I repeat this procedure for million observation and check what is the ratio of each index to occur and it gives me this ratio :

1 -   0.273311
2 -   0.156218 
3 -   0.140228 
4 -   0.156293 
5 -   0.273950 

I ran this simulation multiple times and this ratio stays more or less same. I find this weird. I expected the ratio to be more or less the same for all the indices. If you also observe the 1st and 5th index have same ratio, 2nd and 4th have same ratio.

Instead of taking 5 random variables, I changed it to 10 and it changed the ratio numbers but it still holds the symmetry where 1st and 10th index have the highest value and are same, 2nd and 9th are second highest and same and so on.

Do you know what is going on there?

I don't know what this phenomenon is called so couldn't find appropriate words to google it as well.

This is a reproducible R script with results, if that is of any help.

generate_random_numbers <- function() {
   num = numeric()
   for (i in seq(1000000)) {
     vec = rnorm(5)
     num[i] = which.max(cumsum(vec))
 }
 table(num)/length(num)
}

generate_random_numbers()
#num
#       1        2        3        4        5 
#0.272608 0.156398 0.140701 0.156537 0.273756 

generate_random_numbers()
#num
#       1        2        3        4        5 
#0.273287 0.156035 0.140446 0.156400 0.273832 

Answer 1

What you're observing is a discrete version of an arcsine law for one-dimensional random walks. The third arcsine law states that the point $T$ in the interval $[0,1]$ where a Brownian motion achieves its maximum follows an arcsine distribution: $$P(T\le t)=\frac2\pi\arcsin\sqrt t,$$ so the density of $T$ is $$ f_T(t)=\frac1{\pi\sqrt{t(1-t)}},$$ which has the U-shape that you've observed. A consequence of this arcsine law is that a symmetric random walk is most likely to hit its maximum either very early, or very late over its first $N$ steps.

Answer 2

Suppose we have 5 i.i.d. random variables $X_1, ..., X_5$ with mean zero and with a symmetric distribution, so that $X_1$ and $-X_1$ have the same distribution. I claim that the probability that 1 is the unique max index is the same as the probability that 5 is the unique max index.

Let $E_1$ be the event that index 1 is the unique maximum: \begin{align} X_1 &> X_1 + X_2 \\ X_1 &> X_1 + X_2 + X_3 \\ X_1 &> X_1 + X_2 + X_3 + X_4\\ X_1 &> X_1 + X_2 + X_3 + X_4 + X_5 \end{align} Equivalently: \begin{align} 0 &> X_2 \quad (Eq *)\\ 0 &> X_2 + X_3 \\ 0&> X_2 + X_3 + X_4\\ 0 &> X_2 + X_3 + X_4 + X_5 \end{align} Let $E_5$ be the event that index 5 is the unique maximum: \begin{align} X_1 &<X_1 +X_2+X_3+X_4+X_5\\ X_1 + X_2 &<X_1 +X_2+X_3+X_4+X_5\\ X_1 + X_2 + X_3 &<X_1 +X_2+X_3+X_4+X_5\\ X_1 + X_2 + X_3 + X_4 &<X_1 +X_2+X_3+X_4+X_5 \end{align} equivalently \begin{align} 0&<X_2+X_3+X_4+X_5 \\ 0&<X_3+X_4+X_5\\ 0&<X_4+X_5\\ 0&<X_5 \end{align} Because $\{X_i\}$ have the same distribution as $\{-X_i\}$, $P[E_1]$ is the same as the probability of the following event, call it event $C$ \begin{align} 0&<-X_2+-X_3+-X_4+-X_5\\ 0&<-X_3+-X_4+-X_5\\ 0&<-X_4+-X_5\\ 0&<-X_5 \end{align} equivalently \begin{align} 0&>X_2+X_3+X_4+X_5\\ 0&>X_3+X_4+X_5\\ 0&>X_4+X_5\\ 0&>X_5 \end{align} By relabeling the $X_i$ values we see that $P[C]$ has the same probability as the event specified in (Eq *), so $$ P[E_5]=P[C] = P[E_1]$$