makov chains and property

Question

let $X_t$ for $t \in \{1,2,...,10\}$ be a sequence of independent tosses of a fair coin. We denote heads with 1 and tails with 0. Define the random variable $S_n=\sum_{t=1}^n X_t$ for $n \in \{1,2,...,10\}$. let event $A=S_{10}=8$.

What is the distribution of $X_3|A$ ? (that is the distribution of $X_3$ conditionally on the event $A$)

(b) What is the distribution of $X_4|A$ ?

(c) Conditionally on the event $A$; are $X_3$ and $X_4$ still independent?

(d) What is the distribution of $S_5$ ? (without any conditioning this time).

(e) What is the distribution of $S_5|A$ ?

I Have no idea how to answer this question.any help is appreciated. I know it is to be done using Markov chains but don't understand how. What i attempted was $P(X_{t+1}=1|X_t=1)=a$ and $P(X_{T+1}=0|X_t=0)=b$ then i formed my transition matrix to be $$\begin{pmatrix} a & 1-a \\ b & 1-b\end{pmatrix}.$$

Answer 1

Except in question (d) (which is quite obvious on its own), everything is conditioned on $A$ hence this is as if one had $n=10$ balls in a bag, $w=8$ of them white and $n-w=2$ black, which were drawn one by one, $X_k$ being $1$ if the $k$th ball drawn is white and $0$ otherwise. Thus:

• $X_i=1$ means the ball number $i$ is white. By symmetry this happens with probability $w/n$. Thus, $P[X_i=1\mid A]=w/n$ and $P[X_i=0\mid A]=1-(w/n)$. Likewise for $X_j$ for every $j\ne i$.
• $X_i=X_j=1$ means one draws a white ball, which happens with probability $w/n$, then another white ball, which happens with probability $(w-1)/(n-1)$, hence $P[X_i=X_j=1\mid A]=w(w-1)/(n(n-1))\ne(w/n)^2=P[X_i=1\mid A]P[X_j=1\mid A]$. Conditionally on $A$, $X_i$ and $X_j$ are not independent (and in fact they are negatively associated).
• Unconditionally, $S_n$ is binomial $(n,\frac12)$ and $S_{2n}=S_n+S'_n$ where $(S_n,S'_n)$ is i.i.d. Thus, using $A=[S_{2n}=k]$, one sees that $P[S_n=i\mid A]$ is proportional to $P[S_n=i]P[S_n=k-i]$, that is, to ${n\choose i}{n\choose k-i}$. One recognizes the hypergeometric distribution with parameters $(n,n,k)$, hence, for every $0\leqslant i\leqslant k$, $$ P[S_n=i\mid A]=\frac{{n\choose i}{n\choose k-i}}{{2n\choose k}}. $$