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Let $\{X_t:t=0,1,2,\ldots \}$ be a homogeneous Markov chain with state space $\mathcal{S}=\{1,\ldots,n\}$ and transition matrix $$p_{ij}=\binom{n}{j}\left(\frac{i}{n}\right)^j\left(\frac{n-i}{n}\right)^{n-j}\;\;\text{ for }i,j=1,\ldots,n.$$ I want to show that the process is a Martingale with respect to the natural filtration $\{\mathcal{F}_t\}$. So we compute $$ E[X_{t+1}\mid \mathcal{F}_t](\omega)=E[X_{t+1}\mid \sigma(X_t)](\omega)=E[X_{t+1}\mid X_t]\circ X_t(\omega),$$ where $$E[X_{t+1}\mid X_t](j)=E[X_{t+1}\mid X_t=j],\;\;\;\;\text{ for all } j\in\mathcal{S}.$$ (I'm expecting $E[X_{t+1}\mid X_t=j]$ to be equal to $j$ for the process to be a Matringale). So I computed $$E[X_{t+1}\mid X_t=j]=\frac{E[1_{\{X_t=j\}}X_{t+1}]}{ \mathbb{P}\{X_t=j\}} = \frac{\sum_{i=1}^ni\cdot\mathbb{P}(X_{t+1}=i\mid X_t=j)}{\sum_{i=1}^n p_{ij} \mathbb{P}(X_0=i)}=\frac{\sum_{i=1}^n i\cdot p_{ij}}{\sum_{i=1}^n p_{ij} \mathbb{P}(X_0=i)}$$

And I got stuck on the last part. First of all, is it true that $$E[1_{\{X_t=j\}}X_{t+1}]=\sum_{i=1}^ni\cdot\mathbb{P}(X_{t+1}=i\mid X_t=j) ?$$

If yes how to proceed? I know I still need to replace $p_{ij}$ with it's appropriate value but the computation seems like a mess!

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  • $\begingroup$ Not to nitpick, but you have defined $p_{ij}$ for $i,j=1,\ldots,n$. Should this not be for $i,j=0,\ldots,n$, since $\mathcal S = \{0,\ldots,n\}$? $\endgroup$
    – Math1000
    Nov 22, 2019 at 23:28
  • $\begingroup$ Actually I see a problem there in which $p_{0j}=0$ for all $j>0$ and $p_{00}=1$, so that $0$ is an absorbing state, and similarly $p_{nn}=1$ so that $n$ is an absorbing state as well. $\endgroup$
    – Math1000
    Nov 22, 2019 at 23:31
  • $\begingroup$ @Math1000 it was a typo. Fixed! $\endgroup$
    – UserA
    Nov 22, 2019 at 23:48
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    $\begingroup$ @UserA After the update $\sum_{j}p_{ij}<1$ $\endgroup$
    – user140541
    Nov 22, 2019 at 23:55
  • $\begingroup$ @d.k.o. Indeed because if $X_{n+1}\mid X_n = i$ is $\mathrm{Bin}(n,i/n)$ then we no longer have the probability for $0$ successes. $\endgroup$
    – Math1000
    Nov 23, 2019 at 0:47

1 Answer 1

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I assume that $p_{0j}=1\{j=0\}$ and $p_{nj}=1\{j=n\}$. Then $\mathsf{E}[X_{t+1}\mid X_t=0]=0$, $\mathsf{E}[X_{t+1}\mid X_t=n]=n$, and for $0<i<n$, $$ \mathsf{E}[X_{t+1}\mid X_t=i]=\sum_{j=1}^nj\cdot p_{ij}=n\cdot\frac{i}{n}=i $$ because $X_{t+1}\mid X_t=i\sim \text{Bin}(n,i/n)$.

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  • $\begingroup$ Can you prove this and compare your result to my question? Thank you in advance! $\endgroup$
    – UserA
    Nov 22, 2019 at 23:51
  • $\begingroup$ thank you but you did not answer my question! you got that $E[X_{t+1}\mid X_t=i]=\sum_{j=1}^n j\cdot p_{ij}$ while I got that $E[X_{t+1}\mid X_t=i]=(\sum_{j=1}^n j\cdot p_{ij})/{\Pr\{X_t=i\}}$. What did I do wrong? $\endgroup$
    – UserA
    Nov 23, 2019 at 0:19
  • $\begingroup$ By definition $p_{ij}=\mathsf{P}(X_{t+1}=j\mid X_t=i)$. $\endgroup$
    – user140541
    Nov 23, 2019 at 0:23
  • $\begingroup$ Then $\mathsf{E}[X_{t+1}\mid X_t=i]=\sum_{j=0}^n j\,\mathsf{P}(X_{t+1}=j\mid X_t=i)=\sum_{j=1}^n j\,p_{ij}$. $\endgroup$
    – user140541
    Nov 23, 2019 at 0:42

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