martingales, stochastic processes

Question

Suppose Xn, $n\geqslant0$ is a Markov chain on $\varphi =\left \{ 0,1,2,...,d \right \}$ and $P(x,y)=\frac{\binom{2x}{y}\binom{2d-2x}{d-y}}{\binom{2d}{d}} $. States 0 and d are absorbing states for this chain. Please show that this chain satisfies the equation: $\sum_{y=0}^{d}yP(x,y)=x, x=0,...,d$

I think the equation is the expectation of x, such as Ex(Xi)=x, so I need to take the condition into this form. But i don't know what to do.

Answer 1

No, you don't need expectations. All you need is a straight up combinatorics.

For instance, look at $d=2$ to see what's up here. The numerator and denominator are: $$\sum_{y=0}^2 y \binom{2 x}{y}\binom{4-2 x}{2-y} = 6 x$$ $$\binom{4}{2} = 6$$

So, you get $$\sum_{y=0}^{2}yP(x,y)=x, x=0,1,2$$

UPDATE:

Here's how the numerator looks for $x=1$: $$ 0 \binom{2}{0}\binom{2}{2} = 0$$ $$ \binom{2}{1}\binom{2}{1} = 4$$ $$2 \binom{2}{2}\binom{2}{0} = 2$$ The numerator is the sum $0+4+2=6$, so the equality to holds. You can check $x=0,2$ yourself.