Induction - sum of uniform variables density

Question

In relation to previous question here

How to derive an inductive relationship between the probability density for a sum of $M$ uniform RVs

$P_{X_{1}+...+X_{M}}(u) = \frac{1}{2(n-1)!}\sum^{M}_{k=0}(-1)^{k}\binom{n}{k}(u-k)^{M-1}sgn(u-k) $

and the probability density for the sum of $M-1$ uniforms RV ? In other words how to express $P_{X_{1}+...+X_{M}}(u)$ as a function of $P_{X_{1}+...+X_{M-1}}(u)$ ?

I am using Pascal's equality $\binom{n}{k}=\binom{n}{k-1}+\binom{n-1}{k-1}$

but I get stuck with additional $(u-k)^{M}$ factor for each term of the sum.

Answer 1

Let $Y_{k}=X_{1}+...+X_{k}$ we have $Y_{M}=Y_{M−1}+X_{M}$. The $X_{i}$ are independent and

The probability distribution of the sum of two or more independent random variables is the convolution of their individual distributions.

Hence we have $P[Y_{M}=u]=\sum_{k} P[Y_{M-1}=k]P[X_{M}=u-k]$.