What can we say about the probability distribution function of $n$ independent random variables with different means and variances but the same PDF?
For example, lets say $X_1,X_2,\dots,X_n$ are independent random variables with the following PDFs:
$$
\begin{align}
X_1 & \sim N(μ_1,σ_1) \\
X_2 & \sim N(μ_2,σ_2) \\
& {}\,\vdots \\
X_n & \sim N(μ_n,σ_n)
\end{align}
$$
So they are all Normal random variables with different means and variances and their summation would also have Normal distribution. Now my question is what about other distribution functions?
Best Answer
If $f(x)\,dx$ is a probability distribution with expected value $0$ and variance $1$, and the distribution of $X_i$ is
$$f\left(\dfrac{x-\mu_i}{\sigma_i}\right)\cdot\dfrac{dx}{\sigma_i},$$ and $X_i$ are independent, then certainly the distribution of $X_1+\cdots+X_n$ has expected value $\mu_1+\cdots+\mu_n$ and variance $\sigma_1^2+\cdots+\sigma_n^2$. Also, the higher cumulants would add together in the same way. (The fourth cumulant, for example, is $\mathbb E((X-\mu)^4) - 3(\mathbb E((X-\mu)^2))^2$, and the coefficient $3$ is the only number that makes this functional additive in the sense that the fourth cumulant of a sum of independent random variables is the sum of their fourth cumulants.)
We've tacitly assumed $\sigma_i<\infty$. I think if $\sum_{i=1}^\infty\sigma_i^2=\infty$, then as $n$ grows, the distribution would approach a normal distribution (I'm not recalling the appropriate generalization of the central limit theorem clearly enough to state it precisely.) But what happens for small $n$ is another question, and the answer would depend on what function $f$ is.
I said above that $f(x)\,dx$ has expectation $0$ and variance $1$. But one can also have perfectly good location-scale families in which the expectation, and a fortiori, the variance, do not exist. The most well-known case is the Cauchy distribution. The simplest result there is that $(X_1+\cdots+X_n)/n$ actually has the same Cauchy distribution as $X_1$ if these $n$ variables are i.i.d. It doesn't get narrower. So a lot depends on which function $f$ is.