[Math] Confusion about the range of the sum of i.i.d. random variables

probabilityprobability distributionsrandom variables

Let $X_1, X_2, …X_n$ be independent and uniformly distributed random variables on the interval $[0,1]$. Now suppose I wanted to calculate the probability density function of $Z = X_1 + X_2 + … + X_n$. I think this can be done by $n-1$ successive convolutions, but that's not too important for me right now. My confusion stems from the plot on the bottom which shows the resulting PDF's where $n = 2,3,4,5,6$. Obviously we no longer get a uniformly distributed random variable, but what's puzzling to me is the fact, that the new PDF has range $[0,n]$. This result only makes sense to me if we assume that the $X_i$ actually all have the same range (in this case $0 \leq X_i \leq 1$ for all $i$). Informally, what keeps $X_1$ from being the amount of fuel in a passing car and say $X_2$ the number of passengers in said car?

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You assumed the $X_i$ are uniformly distributed in $[0,1]$ in the first place, so why are you later puzzled that "the $X_i$ all have the same range (in this case $0 \le X_i \le 1$ for all $i$)"?

If you add up $n$ numbers, each in the interval $[0,1]$, then you get a number in the range $[0,n]$.

There is no assumption of units (amount of fuel, number of passengers, etc.) here, but implicitly, writing down $X_1 + \cdots + X_n$ implies that for whatever physical quantity $X_i$ is supposed to model, the sum should make sense. Moreover the physical quantity should follow the probablistic assumption (uniform distribution): number of passengers does not make sense though, since presumably the number of passengers is a nonnegative integer, while $X_i$ takes on any value between $0$ and $1$.

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[Math] Range of i.i.d. normal random variables

I think this paper may help. Someone may have expanded on this in the past 50 years, but it seems like a good place to start:

Tables of range and studentized range, HL Harter

http://www.jstor.org/stable/2237810

Edit: In case you're interested and familiar with R, here is some code that seems to work (for me, at least):

r<-function(x,n){
inner.int<-function(s){
exp(-s^2)*(pnorm(s+x/2)-pnorm(s-x/2))^(n-2)
}
return(n*(n-1)*exp(-x^2/4)/(2*pi)*integrate(inner.int,-Inf,Inf)$value)
}

[Math] Maximum of a sum of random variables

Through the Spitzer identity, it is possible to find some kind of transform of the distribution of $M_n$. Well, not exactly. The Spitzer identity involves the expressions $M^+_n = \max_{0\le k\le n} S_k$, where $S_0 = 0$, $S_k = X_1 + \dots + X_k$, $k\ge 1$. So this translates to the positive part of expression you are interested in. But it is possible to reduce the original question to this one, since, as it was already mentioned in the previous answers, $M_n$ has the same distribution as $X + M_{n-1}^+$, and $X$ is independent of $M_{n-1}^+$.

Going back to the Spitzer identity, it reads $$ \sum_{n=0}^\infty s_n \mathsf E[e^{-\lambda M_n^+}] = \exp\left\{\sum_{n=1}^\infty \frac{s^n}n \mathsf E[e^{-\lambda S_n^+}] \right\}, $$ where $S_n^+ = \max\{S_n,0\}$.

The right-hand side can be calculated more or less explicitly in some cases. For example, if $X\simeq N(0,\sigma^2)$, then $S_n \simeq N(0,\sigma^2 n)$, so $$ \mathsf E[e^{-\lambda S_n}\mathbf{1}_{S_n\ge 0}] = \int_0^\infty e^{-\lambda \sigma x\sqrt{n}} e^{-x^2/2}\frac{dx}{\sqrt{2\pi}} = \sqrt{n} \int_0^\infty e^{-\lambda \sigma y n} e^{-y^2n/2}\frac{dy}{\sqrt{2\pi}}. $$ Therefore, $$ \sum_{n=1}^\infty \frac{s^n}n \mathsf E[e^{-\lambda S_n}\mathbf{1}_{S_n\ge 0}] = \sum_{n=1}^\infty \int_0^\infty \frac{s^n e^{-\lambda\sigma yn} e^{-y^2n/2}}{\sqrt{2\pi n}} dy = \int_0^\infty \operatorname{Li}_{1/2}(s e^{-\lambda\sigma y}e^{-y^2})\frac{dy}{\sqrt{2\pi}}, $$ where $\operatorname{Li}_{s}(x) = \sum_{n=1}^\infty{x^n}n^{-s}$ denotes the polylogarithm. Hence, $$ \sum_{n=1}^\infty \frac{s^n}n \mathsf E[e^{-\lambda S_n^+}] = \sum_{n=1}^\infty \frac{s^n}n \mathsf E[e^{-\lambda S_n}\mathbf{1}_{S_n\ge 0}] + \frac12 \sum_{n=1}^\infty \frac{s^n}{n}\\= \int_0^\infty \operatorname{Li}_{1/2}(s e^{-\lambda\sigma y}e^{-y^2})\frac{dy}{\sqrt{2\pi}} - \frac12\log(1-s). $$

This does not look too neat, but it is not hopeless as it might seem. From here you can calculate expectations (quite easily), variances (not so easily) and get some expressions for higher order moments of $M_n^+$.

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[Math] Range of i.i.d. normal random variables

[Math] Maximum of a sum of random variables

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