# Deriving distribution from conditional distribution

Hi guys I am having problems deriving $$P(X = k)$$ if $$P(X = k|X+Y = n)$$ = $${n}\choose{k}$$ $$\times$$ $$2^{-n}$$

X and Y are i.i.d. random variables with values in $$\mathbb{N_0}$$.

After playing a bit with the formula and using independency of the variables I get to:

$$P(X = k)$$ = $${n}\choose{k}\times$$ $$2^{-n}\times$$ $$\frac{P(X+Y = n)}{P(Y = n-k)}$$

and I could theoretically rewrite $$P(X+Y = n)$$ = $$\sum_{k = 0}^{n}P(X = k) P(Y = n-k)$$, although I am not sure whether the indexing is correct and it does not really help me to get further.

I know from an exercise that at some point I should get to an induction, but I just do not know how. Thank you for any help.

• Hint: the expression you got can be written as $k! \,P(X=k) \,(n-k)! \,P(Y=n-k) = 2^n\, n!\, P(X+Y=n)$ and with a little imagination that can be rewritten $\dfrac{P(X=k)}{\frac{c^k}{k!}d} \dfrac{P(Y=n-k)}{\frac{c^{n-k}}{(n-k)!}d} = \dfrac{P(X+Y=n)}{\frac{(2c)^n}{n!}d^2}$ for suitable $c,d$ – Henry May 15 at 16:32
• Sorry, but could you clarify it a bit more? – Jakub May 15 at 17:32
• Do you know any discrete distributions whose probability mass function involves a factorial in the denominator and a power term in the numerator? – Henry May 15 at 19:22

## 1 Answer

Let $$p_n=P(X=n)$$. Consider $$P(X=n|X+Y=n)=\frac{P(X=n)P(Y=0)}{P(X+Y=n)}=2^{-n},$$ which together with $$P(X+Y=n)=\sum_{k=0}^n p_kp_{n-k}$$ implies $$p_np_0=2^{-n}\sum_{k=0}^n p_kp_{n-k}.\tag1$$ Using $$(1)$$, and induction, you can prove that for all $$n\ge 0$$, that $$p_n=p_0\cdot \frac{(p_1/p_0)^n}{n!}.\tag{2}$$ Summing $$(2)$$ over all $$n\ge 0$$, you get $$\sum_n p_n=1=p_0\sum_n \frac{(p_1/p_0)^n}{n!}=p_0e^{p_1/p_0}\implies p_1=-p_0\log p_0. \tag3$$ Therefore, combining $$(3)$$ and $$(2)$$, $$p_n=p_0\frac{(-\log p_0)^n}{n!}.$$ Therefore, the distribution of $$X$$ is determined up to a free parameter, $$p_0$$. Note that $$X$$ has a Poisson$$(\lambda)$$ distribution with $$\lambda=-\log p_0$$.

In case you have trouble proving $$(2)$$, note that the base cases $$n=0,1$$ are immediate, and for all $$n\ge 2$$, the inductive step is as follows (hidden behind spoilers in case you want to find it yourself):

\begin{align}p_np_0&=2^{-n}\cdot 2p_0p_n+2^{-n}\sum_{k=1}^{n-1} p_kp_{n-k}\\&=2^{-n+1}p_0p_n+2^{-n}\sum_{k=1}^{n-1} p_0^2 \frac{(p_1/p_0)^k}{k!}\frac{(p_1/p_0)^{n-k}}{(n-k)!}\\&=2^{-n+1}p_0p_n+2^{-n}\frac{p_0^2(p_1/p_0)^n}{n!}\sum_{k=1}^{n-1}\binom{n}k\\&=2^{-n+1}p_0p_n+2^{-n}\frac{p_0^2(p_1/p_0)^n}{n!}\cdot (2^n-2), \end{align}which simplifies to $$(2)$$.