S.-T. Yau College Student Mathematics Contests 2022

Problem 1. Let $\left\{X_n\right\}$$\left\{X_n\right\}${X_(n)}\left\{X_{n}\right\}$\left\{X_n\right\}$ be a sequence of Gaussian random variables. Suppose that $X$$X$XX$X$ is a random variable such that $X_n$$X_n$X_(n)X_{n}$X_n$ converges to $X$$X$XX$X$ in distribution as $n\to \mathrm{\infty }$$n\to \mathrm{\infty }$n rarr oon \rightarrow \infty$n\to \mathrm{\infty }$. Show that $X$$X$XX$X$ is also a (possibly degenerate, i.e., variance zero) Gaussian random variable.

Solution: Let $f_n(t)=\mathbb{E}{e}^{it{X}_n}$$f_n(t)=\mathbb{E}{e}^{it{X}_n}$f_(n)(t)=Ee^(itX_(n))f_{n}(t)=\mathbb{E} e^{i t X_{n}}$f_n(t)=\mathbb{E}{e}^{it{X}_n}$ be the characteristic function of $X_n$$X_n$X_(n)X_{n}$X_n$ and $f(t)=\mathbb{E}{e}^{itX}$$f(t)=\mathbb{E}{e}^{itX}$f(t)=Ee^(itX)f(t)=\mathbb{E} e^{i t X}$f(t)=\mathbb{E}{e}^{itX}$ be that of $X$$X$XX$X$. There are real numbers ${\mu }_n$${\mu }_n$mu_(n)\mu_{n}${\mu }_n$ and ${\sigma }_n$${\sigma }_n$sigma_(n)\sigma_{n}${\sigma }_n$ such that $f_n(t)=e^{i{\mu }_{n}t-{\sigma }_n^2{t}^2/2}$$f_n(t)=e^{i{\mu }_{n}t-{\sigma }_n^2{t}^2/2}$f_(n)(t)=e^(imu_(n)t-sigma_(n)^(2)t^(2)//2)f_{n}(t)=e^{i \mu_{n} t-\sigma_{n}^{2} t^{2} / 2}$f_n(t)=e^{i{\mu }_{n}t-{\sigma }_n^2{t}^2/2}$. We have ${|f_n(t)|}^2\to |f(t){|}^2$${\left|f_n(t)\right|}^2\to |f(t){|}^2$|f_(n)(t)|^(2)rarr|f(t)|^(2)\left|f_{n}(t)\right|^{2} \rightarrow|f(t)|^{2}${|f_n(t)|}^2\to |f(t){|}^2$, hence $e^{-{\sigma }_n^2{t}^2}\to |f(t){|}^2$$e^{-{\sigma }_n^2{t}^2}\to |f(t){|}^2$e^(-sigma_(n)^(2)t^(2))rarr|f(t)|^(2)e^{-\sigma_{n}^{2} t^{2}} \rightarrow|f(t)|^{2}$e^{-{\sigma }_n^2{t}^2}\to |f(t){|}^2$ for all $t\in \mathbf{R}$$t\in \mathbf{R}$t inRt \in \mathbf{R}$t\in \mathbf{R}$. Since $f(t)\ne 0$$f(t)\ne 0$f(t)!=0f(t) \neq 0$f(t)\ne 0$ if $t$$t$tt$t$ is close to 0 , we must have ${\sigma }_n^2\to {\sigma }^2$${\sigma }_n^2\to {\sigma }^2$sigma_(n)^(2)rarrsigma^(2)\sigma_{n}^{2} \rightarrow \sigma^{2}${\sigma }_n^2\to {\sigma }^2$ for some $\sigma \in [0,\mathrm{\infty })$$\sigma \in [0,\mathrm{\infty })$sigma in[0,oo)\sigma \in[0, \infty)$\sigma \in [0,\mathrm{\infty })$. Now we have $e^{i{\mu }_{n}t}\to f(t)e^{{\sigma }^2{t}^2}$$e^{i{\mu }_{n}t}\to f(t)e^{{\sigma }^2{t}^2}$e^(imu_(n)t)rarr f(t)e^(sigma^(2)t^(2))e^{i \mu_{n} t} \rightarrow f(t) e^{\sigma^{2} t^{2}}$e^{i{\mu }_{n}t}\to f(t)e^{{\sigma }^2{t}^2}$ for all $t\in \mathbf{R}$$t\in \mathbf{R}$t inRt \in \mathbf{R}$t\in \mathbf{R}$ and by the dominated convergence theorem,

The integral on the right side does not vanish if $t$$t$tt$t$ is close, but not equal to, 0 because the integrand is countinuous and equal to 1 at $s=0$$s=0$s=0s=0$s=0$. On the other hand,

This gives

from which we see that that ${\mu }_n$${\mu }_n$mu_(n)\mu_{n}${\mu }_n$ must converges to a finite number $\mu$$\mu$mu\mu$\mu$. Finally,

and $X$$X$XX$X$ must be a (possibly denegerate) Gaussian random variable.

Problem 2. For two probability measures $\mu$$\mu$mu\mu$\mu$ and $\nu$$\nu$nu\nu$\nu$ on the real line $\mathbf{R}$$\mathbf{R}$R\mathbf{R}$\mathbf{R}$, the total variation distance $\parallel \mu -\nu {\parallel }_{TV}$$\parallel \mu -\nu {\parallel }_{TV}$||mu-nu||_(TV)\|\mu-\nu\|_{T V}$\parallel \mu -\nu {\parallel }_{TV}$ is defined as

where $\mathcal{B}(\mathbf{R})$$\mathcal{B}(\mathbf{R})$B(R)\mathcal{B}(\mathbf{R})$\mathcal{B}(\mathbf{R})$ is the $\sigma$$\sigma$sigma\sigma$\sigma$-algebra of Borel sets on $\mathbf{R}$$\mathbf{R}$R\mathbf{R}$\mathbf{R}$. Let $\mathcal{C}(\mu ,\nu )$$\mathcal{C}(\mu ,\nu )$C(mu,nu)\mathcal{C}(\mu, \nu)$\mathcal{C}(\mu ,\nu )$ be the space of couplings of the probability measures $\mu$$\mu$mu\mu$\mu$ and $\nu$$\nu$nu\nu$\nu$, i.e., the space of ${\mathbf{R}}^2$${\mathbf{R}}^2$R^(2)\mathbf{R}^{2}${\mathbf{R}}^2$ valued random variables $(X,Y)$$(X,Y)$(X,Y)(X, Y)$(X,Y)$ defined on some (not necessarily same) probability space $(\mathrm{\Omega },\mathcal{F},\mathbb{P})$$(\mathrm{\Omega },\mathcal{F},\mathbb{P})$(Omega,F,P)(\Omega, \mathcal{F}, \mathbb{P})$(\mathrm{\Omega },\mathcal{F},\mathbb{P})$ such that the marginal distributions of $X$$X$XX$X$ and $Y$$Y$YY$Y$ are $\mu$$\mu$mu\mu$\mu$ and $\nu$$\nu$nu\nu$\nu$, respectively. Show that

For simplicity you may assume that $\mu$$\mu$mu\mu$\mu$ and $\nu$$\nu$nu\nu$\nu$ are absolutely continuous with respect to the Lebesgue measure on R.

Solution: (1) Let $C\in \mathcal{B}(\mathbf{R})$$C\in \mathcal{B}(\mathbf{R})$C inB(R)C \in \mathcal{B}(\mathbf{R})$C\in \mathcal{B}(\mathbf{R})$ and $(X,Y)\in \mathcal{C}(\mu ,\nu )$$(X,Y)\in \mathcal{C}(\mu ,\nu )$(X,Y)inC(mu,nu)(X, Y) \in \mathcal{C}(\mu, \nu)$(X,Y)\in \mathcal{C}(\mu ,\nu )$. Then

Taking the supremum over $C\in \mathcal{B}(\mathbf{R})$$C\in \mathcal{B}(\mathbf{R})$C inB(R)C \in \mathcal{B}(\mathbf{R})$C\in \mathcal{B}(\mathbf{R})$ and then the infimum over $(X,Y)\in \mathcal{C}(\mu ,\nu )$$(X,Y)\in \mathcal{C}(\mu ,\nu )$(X,Y)inC(mu,nu)(X, Y) \in \mathcal{C}(\mu, \nu)$(X,Y)\in \mathcal{C}(\mu ,\nu )$ we obtain

(2) It is sufficient to a probability measure $\mathbb{P}\in \mathcal{C}(\mu ,\nu )$$\mathbb{P}\in \mathcal{C}(\mu ,\nu )$PinC(mu,nu)\mathbb{P} \in \mathcal{C}(\mu, \nu)$\mathbb{P}\in \mathcal{C}(\mu ,\nu )$ and a set $C\in \mathcal{B}(\mathbf{R})$$C\in \mathcal{B}(\mathbf{R})$C inB(R)C \in \mathcal{B}(\mathbf{R})$C\in \mathcal{B}(\mathbf{R})$ such that for $(X,Y)\in {\mathbf{R}}^2$$(X,Y)\in {\mathbf{R}}^2$(X,Y)inR^(2)(X, Y) \in \mathbf{R}^{2}$(X,Y)\in {\mathbf{R}}^2$ under this probability,

The idea is to construct $\mathbb{P}$$\mathbb{P}$P\mathbb{P}$\mathbb{P}$ such that the probability $\mathbb{P}\{X=Y\}$$\mathbb{P}\{X=Y\}$P{X=Y}\mathbb{P}\{X=Y\}$\mathbb{P}\{X=Y\}$ is the largest possible under the condition that $(X,Y)\in \mathcal{C}(\mu ,\nu )$$(X,Y)\in \mathcal{C}(\mu ,\nu )$(X,Y)inC(mu,nu)(X, Y) \in \mathcal{C}(\mu, \nu)$(X,Y)\in \mathcal{C}(\mu ,\nu )$. Let $m=\mu +\nu$$m=\mu +\nu$m=mu+num=\mu+\nu$m=\mu +\nu$, or just take $m$$m$mm$m$ to be the Lebesgue measure if $\mu$$\mu$mu\mu$\mu$ and $\nu$$\nu$nu\nu$\nu$ are absolutely continuous with respect to $m$$m$mm$m$. We have $\mu =f_1\cdot m$$\mu =f_1\cdot m$mu=f_(1)*m\mu=f_{1} \cdot m$\mu =f_1\cdot m$ and $\nu =f_2\cdot m$$\nu =f_2\cdot m$nu=f_(2)*m\nu=f_{2} \cdot m$\nu =f_2\cdot m$ by the Radon-Nikodym theorem. Let $f=min\{f_1,f_2\}=f_1\wedge f_2$$f=min\left\{f_1,f_2\right\}=f_1\wedge f_2$f=min{f_(1),f_(2)}=f_(1)^^f_(2)f=\min \left\{f_{1}, f_{2}\right\}=f_{1} \wedge f_{2}$f=min\{f_1,f_2\}=f_1\wedge f_2$. Define a probability measure $\mathbb{P}$$\mathbb{P}$P\mathbb{P}$\mathbb{P}$ on ${\mathbf{R}}^2$${\mathbf{R}}^2$R^(2)\mathbf{R}^{2}${\mathbf{R}}^2$ by

Here $a={\int }_{\mathbf{R}}f(z)m(dz)$$a={\int }_{\mathbf{R}} f(z)m(dz)$a=int_(R)f(z)m(dz)a=\int_{\mathbf{R}} f(z) m(d z)$a={\int }_{\mathbf{R}}f(z)m(dz)$ and we assume that $a<1$$a<1$a < 1a<1$a<1$; otherwise $a=1$$a=1$a=1a=1$a=1$ and $f_1=f_2$$f_1=f_2$f_(1)=f_(2)f_{1}=f_{2}$f_1=f_2$, and the case is trivial. Note that the first part is the product measure of $(f_1-f)\cdot m$$\left(f_1-f\right)\cdot m$(f_(1)-f)*m\left(f_{1}-f\right) \cdot m$(f_1-f)\cdot m$ and $(f_2-f)\cdot m$$\left(f_2-f\right)\cdot m$(f_(2)-f)*m\left(f_{2}-f\right) \cdot m$(f_2-f)\cdot m$ ) (up to a constant) and the second part is the probability measure $f\cdot m$$f\cdot m$f*mf \cdot m$f\cdot m$ on the diagonal (identified with $\mathbf{R}$$\mathbf{R}$R\mathbf{R}$\mathbf{R}$ ) of ${\mathbf{R}}^2$${\mathbf{R}}^2$R^(2)\mathbf{R}^{2}${\mathbf{R}}^2$. We have

Similarly $\mathbb{P}\{Y\in B\}=\nu (B)$$\mathbb{P}\{Y\in B\}=\nu (B)$P{Y in B}=nu(B)\mathbb{P}\{Y \in B\}=\nu(B)$\mathbb{P}\{Y\in B\}=\nu (B)$, hence $(X,Y)\in \mathcal{C}(\mu ,\nu )$$(X,Y)\in \mathcal{C}(\mu ,\nu )$(X,Y)inC(mu,nu)(X, Y) \in \mathcal{C}(\mu, \nu)$(X,Y)\in \mathcal{C}(\mu ,\nu )$. On the other hand,

If we choose $C=\{f_1>f_2\}$$C=\left\{f_1>f_2\right\}$C={f_(1) > f_(2)}C=\left\{f_{1}>f_{2}\right\}$C=\{f_1>f_2\}$, then

This shows that $\mu (C)-\nu (C)=\mathbb{P}\{X\ne Y\}$$\mu (C)-\nu (C)=\mathbb{P}\{X\ne Y\}$mu(C)-nu(C)=P{X!=Y}\mu(C)-\nu(C)=\mathbb{P}\{X \neq Y\}$\mu (C)-\nu (C)=\mathbb{P}\{X\ne Y\}$.

Problem 3. We throw a fair die repeatedly and independently. Let ${\tau }_{11}$${\tau }_{11}$tau_(11)\tau_{11}${\tau }_{11}$ be the first time the pattern 11 (two consecutive 1 's) appears and ${\tau }_{12}$${\tau }_{12}$tau_(12)\tau_{12}${\tau }_{12}$ the first time the pattern 12 ( 1 followed by 2 ) appears.

(a) Calculate the expected value $\mathbb{E}{\tau }_{11}$$\mathbb{E}{\tau }_{11}$Etau_(11)\mathbb{E} \tau_{11}$\mathbb{E}{\tau }_{11}$.

(b) Which is larger, $\mathbb{E}{\tau }_{11}$$\mathbb{E}{\tau }_{11}$Etau_(11)\mathbb{E} \tau_{11}$\mathbb{E}{\tau }_{11}$ or $\mathbb{E}{\tau }_{12}$$\mathbb{E}{\tau }_{12}$Etau_(12)\mathbb{E} \tau_{12}$\mathbb{E}{\tau }_{12}$ ? It is sufficient to give an intuitive argument to justify your answer. You can also calculate $\mathbb{E}{\tau }_{12}$$\mathbb{E}{\tau }_{12}$Etau_(12)\mathbb{E} \tau_{12}$\mathbb{E}{\tau }_{12}$ if you wish.

(a) Let ${\tau }_1$${\tau }_1$tau_(1)\tau_{1}${\tau }_1$ be the first time the digit 1 appears. At this time, if the next result is 1 , then ${\tau }_{11}={\tau }_1+1$${\tau }_{11}={\tau }_1+1$tau_(11)=tau_(1)+1\tau_{11}=\tau_{1}+1${\tau }_{11}={\tau }_1+1$; if the next result is not 1 , then the time is ${\tau }_1+1$${\tau }_1+1$tau_(1)+1\tau_{1}+1${\tau }_1+1$ and we have to start all over again. This means

Solving for $\mathbb{E}{\tau }_{11}$$\mathbb{E}{\tau }_{11}$Etau_(11)\mathbb{E} \tau_{11}$\mathbb{E}{\tau }_{11}$ we have $\mathbb{E}{\tau }_{11}=6(\mathbb{E}{\tau }_1+1)$$\mathbb{E}{\tau }_{11}=6\left(\mathbb{E}{\tau }_1+1\right)$Etau_(11)=6(Etau_(1)+1)\mathbb{E} \tau_{11}=6\left(\mathbb{E} \tau_{1}+1\right)$\mathbb{E}{\tau }_{11}=6(\mathbb{E}{\tau }_1+1)$. We need to calculate $\mathbb{E}{\tau }_1$$\mathbb{E}{\tau }_1$Etau_(1)\mathbb{E} \tau_{1}$\mathbb{E}{\tau }_1$. The set $\{{\tau }_1\ge n\}$$\left\{{\tau }_1\ge n\right\}${tau_(1) >= n}\left\{\tau_{1} \geq n\right\}$\{{\tau }_1\ge n\}$ is the event that that none of the first $n-1$$n-1$n-1n-1$n-1$ results is 1 , hence $\mp \{{\tau }_1\ge n\}=(5/6{)}^{n-1}$$\mp \left\{{\tau }_1\ge n\right\}=(5/6{)}^{n-1}$∓{tau_(1) >= n}=(5//6)^(n-1)\mp\left\{\tau_{1} \geq n\right\}=(5 / 6)^{n-1}$\mp \{{\tau }_1\ge n\}=(5/6{)}^{n-1}$ and

It follows that $\mathbb{E}{\tau }_{11}=6(6+1)=42$$\mathbb{E}{\tau }_{11}=6(6+1)=42$Etau_(11)=6(6+1)=42\mathbb{E} \tau_{11}=6(6+1)=42$\mathbb{E}{\tau }_{11}=6(6+1)=42$.

(b) For either 11 or 12 to occur, we have to wait until the first 1 occurs. After that, if we want 11, the next digit needs to be 1 ; otherwise we have to start all over again (i.e., waiting for the next 1 ). But if we want 12 , the next digit needs to be 2 ; otherwise, we have to start all over again only if the next digit is 3 to 6 because if the next digit is 1 , we have already have a start on the pattern 12. It follows that the pattern 12 has a slight advantage to occur earlier than 11. Thus we have $\mathbb{E}{\tau }_{12}\le \mathbb{E}{\tau }_{11}$$\mathbb{E}{\tau }_{12}\le \mathbb{E}{\tau }_{11}$Etau_(12) <= Etau_(11)\mathbb{E} \tau_{12} \leq \mathbb{E} \tau_{11}$\mathbb{E}{\tau }_{12}\le \mathbb{E}{\tau }_{11}$.

We can also calculate $\mathbb{E}{\tau }_{12}$$\mathbb{E}{\tau }_{12}$Etau_(12)\mathbb{E} \tau_{12}$\mathbb{E}{\tau }_{12}$ directly. Let ${\tau }_1$${\tau }_1$tau_(1)\tau_{1}${\tau }_1$ be as before and let $\sigma$$\sigma$sigma\sigma$\sigma$ be the first time a digit not equal to 1 appears. After ${\tau }_1$${\tau }_1$tau_(1)\tau_{1}${\tau }_1$ we wait until the first time a digit not equal to 1 appears. With probability $1/5$$1/5$1//51 / 5$1/5$ this digit is 2 ; with probability $4/5$$4/5$4//54 / 5$4/5$ this probability is not 2 , then we have to start over again. This means that

Hence $\mathbb{E}{\tau }_{12}=5\mathbb{E}({\tau }_1+\sigma )$$\mathbb{E}{\tau }_{12}=5\mathbb{E}\left({\tau }_1+\sigma \right)$Etau_(12)=5E(tau_(1)+sigma)\mathbb{E} \tau_{12}=5 \mathbb{E}\left(\tau_{1}+\sigma\right)$\mathbb{E}{\tau }_{12}=5\mathbb{E}({\tau }_1+\sigma )$. We have seen $\mathbb{E}{\tau }_1=6$$\mathbb{E}{\tau }_1=6$Etau_(1)=6\mathbb{E} \tau_{1}=6$\mathbb{E}{\tau }_1=6$. On the other hand, $\{\sigma \ge n\}$$\{\sigma \ge n\}${sigma >= n}\{\sigma \geq n\}$\{\sigma \ge n\}$ is the event that the first $n-1$$n-1$n-1n-1$n-1$ digits are 1 , hence $\mp \{\sigma \ge n\}=(1/6{)}^{n-1}$$\mp \{\sigma \ge n\}=(1/6{)}^{n-1}$∓{sigma >= n}=(1//6)^(n-1)\mp\{\sigma \geq n\}=(1 / 6)^{n-1}$\mp \{\sigma \ge n\}=(1/6{)}^{n-1}$ and $\mathbb{E}\sigma =6/5$$\mathbb{E}\sigma =6/5$Esigma=6//5\mathbb{E} \sigma=6 / 5$\mathbb{E}\sigma =6/5$. It follows that

Problem 4. Let $\left\{X_n\right\}$$\left\{X_n\right\}${X_(n)}\left\{X_{n}\right\}$\left\{X_n\right\}$ be a Markov chain on a discrete state space $S$$S$SS$S$ with transition function $p(x,y),x,y\in S$$p(x,y),x,y\in S$p(x,y),x,y in Sp(x, y), x, y \in S$p(x,y),x,y\in S$. Suppose that there is a state $y_0\in S$$y_0\in S$y_(0)in Sy_{0} \in S$y_0\in S$ and a positive number $\theta$$\theta$theta\theta$\theta$ such that $p(x,y_0)\ge \theta$$p\left(x,y_0\right)\ge \theta$p(x,y_(0)) >= thetap\left(x, y_{0}\right) \geq \theta$p(x,y_0)\ge \theta$ for all $x\in S$$x\in S$x in Sx \in S$x\in S$.

(a) Show that is a positive constant $\lambda <1$$\lambda <1$lambda < 1\lambda<1$\lambda <1$ such that for any two initial distribution $\mu$$\mu$mu\mu$\mu$ and $\nu$$\nu$nu\nu$\nu$,

(b) Show that the Markov chain has a unique stationary distribution $\pi$$\pi$pi\pi$\pi$ and

(a) Let $\theta =min\{p(x,y_0):x\in S\}$$\theta =min\left\{p\left(x,y_0\right):x\in S\right\}$theta=min{p(x,y_(0)):x in S}\theta=\min \left\{p\left(x, y_{0}\right): x \in S\right\}$\theta =min\{p(x,y_0):x\in S\}$. Then $0<\theta \le 1$$0<\theta \le 1$0 < theta <= 10<\theta \leq 1$0<\theta \le 1$. For any two probability meausres $\mu$$\mu$mu\mu$\mu$ and $\nu$$\nu$nu\nu$\nu$ on the state space $S$$S$SS$S$, we have

For the term $y=y_0$$y=y_0$y=y_(0)y=y_{0}$y=y_0$ we can replace $p(x,y_0)$$p\left(x,y_0\right)$p(x,y_(0))p\left(x, y_{0}\right)$p(x,y_0)$ by $p(x,y_0)-\theta$$p\left(x,y_0\right)-\theta$p(x,y_(0))-thetap\left(x, y_{0}\right)-\theta$p(x,y_0)-\theta$ because $\sum _{x\in S}\{\mu (x)-\nu (x)\}=1-1=0$$\sum _{x\in S} \{\mu (x)-\nu (x)\}=1-1=0$sum_(x in S){mu(x)-nu(x)}=1-1=0\sum_{x \in S}\{\mu(x)-\nu(x)\}=1-1=0$\sum _{x\in S}\{\mu (x)-\nu (x)\}=1-1=0$. After this replacement, we take the absolute value of every term and exchange the order of summation. Using the fact that $p(x,y_0)-\theta \ge 0$$p\left(x,y_0\right)-\theta \ge 0$p(x,y_(0))-theta >= 0p\left(x, y_{0}\right)-\theta \geq 0$p(x,y_0)-\theta \ge 0$ we have

The first sum on the right side is $1-\theta =\lambda <1$$1-\theta =\lambda <1$1-theta=lambda < 11-\theta=\lambda<1$1-\theta =\lambda <1$. It follows that

(b) Let ${\mu }_n(x)={\mathbb{P}}_{\mu }\{X_n=x\}$${\mu }_n(x)={\mathbb{P}}_{\mu }\left\{X_n=x\right\}$mu_(n)(x)=P_(mu){X_(n)=x}\mu_{n}(x)=\mathbb{P}_{\mu}\left\{X_{n}=x\right\}${\mu }_n(x)={\mathbb{P}}_{\mu }\{X_n=x\}$. Then ${\mu }_{n+1}={\mathbb{P}}_{{\mu }_n}\{X_1=x\}$${\mu }_{n+1}={\mathbb{P}}_{{\mu }_n}\left\{X_1=x\right\}$mu_(n+1)=P_(mu_(n)){X_(1)=x}\mu_{n+1}=\mathbb{P}_{\mu_{n}}\left\{X_{1}=x\right\}${\mu }_{n+1}={\mathbb{P}}_{{\mu }_n}\{X_1=x\}$ and ${\mu }_n={\mathbb{P}}_{{\mu }_{n-1}}\{X_1=x\}$${\mu }_n={\mathbb{P}}_{{\mu }_{n-1}}\left\{X_1=x\right\}$mu_(n)=P_(mu_(n-1)){X_(1)=x}\mu_{n}=\mathbb{P}_{\mu_{n-1}}\left\{X_{1}=x\right\}${\mu }_n={\mathbb{P}}_{{\mu }_{n-1}}\{X_1=x\}$. By (a),

It follows that