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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?

taken from the german Wikipedia article on the uniform dist.

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    $\begingroup$ If someone says $Y$ is a uniformly distributed random variable on the interval $[0,1]$, it implies $0\leq Y \leq1$. You were told all the $X_i$ are variables of that type; that's how you know $0\leq X_i \leq1$. You already have an answer that explains this in more detail. $\endgroup$
    – David K
    Commented Jan 16, 2016 at 22:19
  • $\begingroup$ Short answer: the final "i.d." in the "i.i.d." of your title stands for "identically distributed". So yes, the $X_i$ have the same range (and it is $[0,1]$). I don't know about fuel, but the number of passengers (or any other integer-valued variable) cannot be uniformly distributed on $[0,1]$. $\endgroup$
    – Marc van Leeuwen
    Commented Jan 17, 2016 at 5:35
  • $\begingroup$ I guess the real source of my confusion is that I still don't really understand the way that random variables are used in probability theory. I always thought of them as just plain old functions (as long as the fulfill the measurability criterion) which get evaluated at random points depending on the underlying probability space. I assumed that the fact $X$ is uniformly distributed just means that $X$ gets "evaluated uniformly"... $\endgroup$
    – jazzinsilhouette
    Commented Jan 17, 2016 at 6:06

3 Answers 3

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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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  • $\begingroup$ The way I understand random variables is that they need to be measurable functions from a probability space $(\Omega, E, \mathbb{P})$ into a measurable space $(\Sigma, F)$, e.g. $X: \Omega \to \Sigma$. In this case I don't see why $\Sigma$ needs to equal $[0,1]$. For example, what if $X_{1}(x) = 3x$? The only randomness involved is the selection of points $w \in \Omega$ at which $X_{1}$ is evaluated. Why is it not possible to "amplify" the input value $\omega$, so that for example $\Sigma = [0,3]$? $\endgroup$
    – jazzinsilhouette
    Commented Jan 16, 2016 at 21:07
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    $\begingroup$ @jazzinsilhouette Strictly speaking $\Sigma$ does not have to be $[0,1]$. However, if $X$ is uniform on $[0,1]$ then for all $\omega$ not in some set $N$ of probability zero, $X(\omega) \in [0,1]$. This is just the definition of a uniform random variable. $\endgroup$
    – Ian
    Commented Jan 16, 2016 at 21:12
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    $\begingroup$ @jazzinsilhouette That is fine, but then $X_1$ is no longer uniformly distributed. It seems you are considering arbitrary random variables $X_1,\ldots, X_n$. By convention, we usually consider real random variables, so $X_i : \Omega \to \mathbb{R}$ for all $i$. They all have the same "range" $\mathbb{R}$, and you can "amplify" and modify all you want (within the reals). $\endgroup$
    – angryavian
    Commented Jan 16, 2016 at 21:13
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It makes complete sense.

Ask yourself this: What is the smallest possible value of $Z$? It is $0$.

What is the largest possible value? This occurs if all $X_i$ happen to be $1$, hence the largest value of $Z$ is $n$. So you can deduce that the range of $Z$ is $[0,n]$.

You stated

This result only makes sense to me if we assume that the $X_i$ actually all have the same range

They do. You told us each $X_i$ follows a unif$[0,1]$.

For "large" $n$, the sum of iid random variables converges to a normal distribution by the CLT.

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If your random variables have different units (e.g. number of passengers, mileage, amount of fuel in the tank), then the addition (+) is not defined, and $X_1+X_2$ makes no sense.

To illustrate why addition of two uniform variables is not uniform, lets have a look at the sum of two dices

Dice #1: $$p(d_1=1)=\frac{1}{6}$$ $$p(d_1=2)=\frac{1}{6}$$ $$p(d_1=3)=\frac{1}{6}$$ $$p(d_1=4)=\frac{1}{6}$$ $$p(d_1=5)=\frac{1}{6}$$ $$p(d_1=6)=\frac{1}{6}$$

Same for Dice #2

But, have a look at the sum $$p(d_1+d_2=2)=\frac{1}{36}$$ $$p(d_1+d_2=3)=\frac{2}{36}$$ Already we see that the resulting distribution is not uniform

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