$B_{i}^{n}(x)={{n}\choose{i}}x^i(1-x)^{n-i}$, prove that $B_{i}^{n}(cu)=\sum\limits_{j=0}^{n}B_{i}^{j}(c)B_{j}^{n}(u)$

Question

$B_{i}^{n}(x)={{n}\choose{i}}x^i(1-x)^{n-i}$, prove that $B_{i}^{n}(cu)=\sum\limits_{j=0}^{n}B_{i}^{j}(c)B_{j}^{n}(u)$

I tried to to solve it from the right side: $\sum\limits_{j=0}^{n}B_{i}^{j}(c)B_{j}^{n}(u)=\sum\limits_{j=0}^{n}{{j}\choose{i}}c^i(1-c)^{j-i}{{n}\choose{j}}u^j(1-u)^{n-j}$. Since ${{j}\choose{i}}$ is $0$ for $j<i$, then the sum is $\sum\limits_{j=i}^{n}B_{i}^{j}(c)B_{j}^{n}(u)=\sum\limits_{j=0}^{n-i}B_{i}^{i+j}(c)B_{i+j}^{n}(u)=\sum\limits_{j=0}^{n-i}{{i+j}\choose{i}}c^{i}(1-c)^{j}{{n}\choose{i+j}}u^{j+i}(1-u)^{n-j-i}$.

But I'm stuck...

Answer 1

Simplifying your first attempt yields $$(1-c)^{-i} c^i \sum_{j=0}^n {n\choose j} {j\choose i} (1-c)^j u^j (1-u)^{n-j} .$$

To evaluate the sum, observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that $$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$ i.e. the product of the two generating functions is the generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$ In the present case we have $$A(z) = \sum_{q\ge 0} {q\choose i} (1-c)^q u^q \frac{z^q}{q!} = \frac{1}{i!} (1-c)^i u^i z^i \exp((1-c)uz)$$ and $$B(z) = \sum_{q\ge 0} (1-u)^q \frac{z^q}{q!} = \exp((1-u)z).$$ It follows that the sum is given by $$n![z^n] A(z) B(z) = n! [z^n] \frac{1}{i!} (1-c)^i u^i z^i \exp((1-cu)z).$$ This is $$ n! \frac{1}{i!} (1-c)^i u^i [z^{n-i}] \exp((1-cu)z) = n!\frac{1}{i!} (1-c)^i u^i \frac{(1-cu)^{n-i}}{(n-i)!}.$$ Therefore the value of the convolution is $$(1-c)^{-i} c^i (1-c)^i u^i {n\choose i} (1-cu)^{n-i} = {n\choose i} (cu)^i (1-cu)^{n-i}.$$ There are several errors in the problem statement as of Thu Dec 12 22:04:29 CET 2013. (Fixed by OP as of Thu Dec 12 22:56:42 CET 2013.)

The same trick is also used at this MSE calculation, which shows a nice variety of approaches by several different MSE users.

Post Scriptum. The evaluation of $$A(z) = \sum_{q\ge 0}{q\choose i} \frac{z^q}{q!}$$ is done as follows. $$A(z) = \sum_{q\ge i}{q\choose i} \frac{z^q}{q!} = \sum_{m\ge 0} {m+i\choose i} \frac{z^{m+i}}{(m+i)!} \\= \frac{1}{i!} z^i \sum_{m\ge 0} \frac{(m+i)!}{m!} \frac{z^m}{(m+i)!} = \frac{1}{i!} z^i \sum_{m\ge 0} \frac{z^m}{m!} = \frac{1}{i!} z^i \exp(z).$$

Post Scriptum II. Here is a related MSE calculation II that uses the same technique.

Answer 2

This has a nice proof if you interpret it in terms of probabilities. When $0<x<1$, $B^n_i(x)$ is the probability of $i$ successes from $n$ independent events if each event has a probability $x$ of success.

Now, suppose that in order to succeed, you need to pass two (independent) stages: step 1 has a probability $u$ of success, and step 2 a probability $c$ of success. So the chance of $i$ successes from $n$ events is $B^n_i(cu)$. On the other hand, for each possible $j$, we can compute the probabilities of having $j$ successes at stage $1$, and of those $j$ stage 1 successes, having $i$ of them be successes at stage $2$. For a fixed $j$, the probability of this happening is $B^n_j(u)B^j_i(c)$. In any given instance of $i$ total successes from $n$ events, this must happen for exactly one $j$ with $0\leq j \leq n$, hence by adding we obtain the identity $$B^n_i(cu) = \sum_{j=0}^n B^j_i(c)B^n_j(u)$$ as required.

If you now need the identity to hold for any $c,u$ and not just ones between $0$ and $1$, you can appeal to the fact that you have a polynomial identity (in two variables $c$ and $u$) which holds in the unit square and thus the polynomials are in fact identical.