The best and simpler way to derive the distribution of the sum of two independent Uniform random variates is geometrical requiring working out some area calculations in 2-D.
However, the derivation of the distribution of $\sum_{i=1}^{n}X_i$ for $n>2$ with each $X_i$ independently distributed as $U(0,1)$ is generally tedious. Geometrically it is difficult to visualise for higher values of $n$. However, a convolution approach would be used to find them (that I will use here kind of recursively).
I start assuming you know the distribution of $A=X_1+X_2$ given by the below pdf:
$$f_A(a) = \begin{cases} a & \text{if $0 \le a \le 1$}\\ 2-a & \text{if $1 \le a \le 2$}\\ 0 & \text{elsewhere}\end{cases}$$
For $n=3$, define $B=X_1+X_2+X_3=A+X_3$. Note $0\le B\le3$. Now, by convolution of pdfs, the pdf of B: $f_B(b)=\int_{-\infty}^\infty f_{X_3}(x_3)f_A(b-x_3)\,dx_3$
Note-1: $f_A(b-x_3)=b-x_3$ for $0\le b-x_3\le1$, i.e., $b-1\le x_3\le b$; Also, $0 \le x_3 \le 1$. Combining these two gives $\mathbb{max}(b-1,0) \le x_3\le \mathbb{min}(b,1)$
Note-2: $f_A(b-x_3)=2-b+x_3$ for $1\le b-x_3\le2$, i.e., $b-2\le x_3\le b-1$; Also, $0 \le x_3 \le 1$. Combining these two gives $\mathbb{max}(b-2,0) \le x_3\le \mathbb{min}(b-1,1)$
Looking at the bounds of $x_3$, it is reasonable to break the range of $b$ (i.e.,[$0,3$]) into $0\le b\le1$, $1\le b\le2$ and $2\le b\le3$ and considering the form of the pdf of B within each range separately.
Case: $0\le b\le1$: Range in Note-1 reduces to $0\le x_3\le b$; while that in Note-2 doesn't reduce to a feasible bound for $x_3$. Thus the pdf of $B$ reduces to
$f_B(b)=\int_0^b (b-x_3)\,dx_3=\frac{b^2}{2}$
Case: $1\le b\le2$: Range in Note-1 reduces to $b-1\le x_3\le 1$; while that in Note-2 reduces to $0\le x_3\le b-1$. Thus the pdf of $B$ reduces to
$f_B(b)=\int_{b-1}^1 (b-x_3)\,dx_3+\int_0^{b-1}(2-b+x_3)\,dx_3=\frac{-2b^2+6b-3}{2}$
Case: $2\le b\le3$: Range in Note-1 doesn't reduce to a feasible bound for $x_3$; while that in Note-2 reduces to $b-2\le x_3\le 1$. Thus the pdf of $B$ reduces to
$f_B(b)=\int_{b-2}^1(2-b+x_3)\,dx_3=\frac{(3-b)^2}{2}$
Can you now try similar logic for $n=4$ and $n=5$?
Although it may not be readily intuitive that the general form of the pdf of the sum of Uniform variates would tend to follow approximately Normal for large n, the pdf is a piecewise polynomial function of degree n-1. And, if you plot this function you'll end up with a plot of close to a pdf of Normal distribution (I tried in R) and it indeed goes to show how powerful the CLT is!
$\{U_i\}_{i=1}^n$ n iid uniform distributed variables, and $Z=\sum_{i=1}^{n}{U_i}$
The moment generating function of $Z$ is $$\phi_Z(u)=E(e^{uZ})$$
$$\phi_Z(u)=E(e^{u\sum_{i=1}^{n}{U_i}})$$
$$=E(\prod_{i=1}^{n}e^{u{U_i}})$$
By independence , we have
$$E(\prod_{i=1}^{n}e^{u{U_i}})=\prod_{i=1}^{n}E(e^{u{U_i}})$$
Because the $U_i$'s have the same distribution , we have
$$\phi_Z(u)=E(e^{u{U_1}})^n$$ or
$$\phi_Z(u)=\phi_{U_1}(u)^n$$ where $\phi_{U_1}$ is the moment generating function of a uniform variable, which is defined by
$$\phi_{U_1}(u)=\frac{e^u-1}{u}$$ if $u$ non nil, $1$ otherwise
Best Answer
This is correct. Let $X_1,\ldots,X_n$ be independent and uniformly distributed on $[-1, 1]$, with $X = X_1 + \ldots + X_n$, so $X$ has the modified Irwin-Hall distribution you define. Pick a tiny interval $[x, x+\epsilon] \subseteq [-1, 1]$. Since each $\frac{X_i + 1}{2}$ is uniform on $[0, 1]$, their sum $\frac{X+n}{2}$ has the usual Irwin-Hall distribution. The probability that $X$ falls in $[x, x+\epsilon]$ is equal to the probability that $\frac{X+n}{2}$ falls in $[\frac{x+n}{2}, \frac{x+\epsilon+n}{2}]$. For small $\epsilon$, this is approximately the length of the interval $[\frac{x+n}{2}, \frac{x+\epsilon+n}{2}]$ times the density $f$ of the usual Irwin-Hall distribution at the left endpoint. The result is a probability of about $\frac{\epsilon}{2} f(\frac{x+n}{2})$.
Given the probability that $X$ lies on the interval $[x, x+\epsilon]$, we can now obtain the density of $X$ at $x$ by dividing by $\epsilon$. The answer won't be exactly correct for any positive $\epsilon$ but the limit as $\epsilon \rightarrow 0$ is the true density. Thus we compute that indeed, the density of the modified distribution is $\frac{1}{2} f(\frac{x+n}{2})$.