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Probability paper2022

 
  • Oct 07, 2022
  1. Let $Z_n,n⩾1$, be random variables and $c∈ℝ$. Show that $Z_n→c$ in probability if and only if $Z_n→c$ in distribution.
  2. Fix $λ∈(0,∞)$. For each $r∈(0,∞)$, consider a random variable $X_r$ with probability density function $$ f_{r,λ}(x)=\begin{cases}\frac{1}{Γ(r)}x^{r-1}λ^r e^{-λ x},&x∈(0,∞),\\0,&\text { otherwise. }\end{cases} $$ Recall that this means that $X_r$ has the Gamma distribution with shape parameter $r$ and rate parameter $λ$.
    1. Carefully derive the moment generating function of $X_r$.
    2. Show that $X_r/r$ converges in distribution as $r→∞$. Does this convergence hold in probability?
      [You may use standard theorems about moment generating functions without proof.]
    1. Define the Poisson process $\left(N_t,t⩾0\right)$ of rate $λ∈(0,∞)$ in terms of properties of its increments over disjoint time intervals.
    2. Show that the first arrival time $T_1=\inf\left\{t⩾0:N_t=1\right\}$ is exponentially distributed with parameter $λ∈(0,∞)$.
    3. Show that $T_n=\inf\left\{t⩾0:N_t=n\right\}$ has a Gamma distribution for all $n⩾1$.
      [If you use the inter-arrival time definition of the Poisson process, you are expected to prove that it is equivalent to the definition given in (i).]
    4. Let $R_n,n⩾1$, be independent Gamma-distributed random variables. Let $R_n$ have shape parameter $n$ and rate parameter $λ$.
      Let $Y_t=\#\left\{n⩾1:R_n⩽t\right\},t⩾0$. Show that $Y_t$ is not a Poisson process with rate $λ$, but does satisfy $ℙ\left(Y_t<∞\right.$ for all $\left.t⩾0\right)=1$.
      [Hint: Let $B_n=1_{\left\{R_n⩽t\right\}}$ and write $Y_t=\sum_{n⩾1}B_n$.]
  1. "⇒": Let $x<c$. Then $ℙ\left(Z_n≤x\right)≤ℙ\left(\left|Z_n-c\right|≥c-x\right)→0$. Let $x>c$. Then $ℙ\left(Z_n≤x\right)=1-ℙ\left(Z_n>x\right)≥1-ℙ\left(\left|Z_n-c\right|>x-c\right)→1$. As the limit is the cdf of the constant $c$, which is discontinuous at $x=c$, convergence in distribution holds.
    "⇐": Let $ε>0$. Then $ℙ\left(\left|Z_n-c\right|>ε\right)≤ℙ\left(Z_n≤c-ε\right)+1-ℙ\left(Z_n≤c+ε\right)→0+1-1=0$. Hence convergence in probability holds.
    1. For all $t<λ$, the pdf $f_{r,λ}$ integrates to 1, hence\begin{aligned} M_{X_{r}}(t)=𝔼\left[e^{t X_{r}}\right] &=∫_{0}^∞e^{t x} \frac{1}{Γ(r)} x^{r-1} λ^{r} e^{-λ x} d x \\ &=\frac{λ^{r}}{(λ-t)^{r}} ∫_{0}^∞f_{r, λ-t}(x) d x=\left(\frac{λ}{λ-t}\right)^{r}\end{aligned}
    2. For all $t<rλ$, we have$$M_{X_{r}/r}(t)=M_{X_{r}}(t/r)=\left(\frac{λ}{λ-\frac{t}{r}}\right)^{r}=\frac{1}{\left(1-\frac{t/λ}{r}\right)^{r}}→e^{t/λ}$$In particular, this holds for all $t∈ℝ$ and $r$ sufficiently large. Since the limit is the mgf of the constant $c=1/λ$, the convergence theorem for mgfs yields $X_r/r→c=1/λ$ in distribution, and by (a) also in probability. Alternatively, convergence in probability can be established directly using Chebyshev's inequality, but require also the calculation of the mean and variance of $X_r$.
    1. $\left(N_t,t≥0\right)$ is called $\operatorname{PP}(λ)$ if (1) $N_0=0,(2)N_t-N_s∼\operatorname{Poi}((t-s)λ)$ for all $0≤s<t$, and (3) for all $n≥2$ and $0≤s_1<t_1≤s_2<t_2≤⋯≤s_n<t_n$, the variables $N_{t_j}-N_{s_j},1≤j≤n$, are independent.
    2. $ℙ\left(T_1>t\right)=ℙ\left(N_t-N_0=0\right)=e^{-λ t}$ by (1)-(2), for all $t≥0$. Hence $T_1$ is exponentially distributed with parameter $λ$.
    3. Similarly, for all $t≥0$ $$ ℙ\left(T_n>t\right)=ℙ\left(N_t≤n-1\right)=e^{-λ t}\sum_{k=0}^{n-1}\frac{(λ t)^k}{k !}. $$ Differentiation yields $-f_{T_n}(t)$, hence $$ f_{T_n}(t)=λ e^{-λ t}\sum_{j=0}^{n-1}\frac{(λ t)^j}{j !}-e^{-λ t}\sum_{k=1}^{n-1}\frac{λ^k t^{k-1}}{(k-1)!}=e^{-λ t}λ^n t^{n-1}\frac{1}{(n-1)!}=f_{n,λ}(t), $$ the pdf of a Gamma distribution with shape parameter $n$ and rate parameter $λ$.
    4. First note that the first arrival time of $\left(Y_t,t≥0\right)$ is $\inf\left\{R_n,n≥1\right\}≤R_1$, and this is strictly smaller than $R_1$ with positive probability since $ℙ\left(R_2<R_1\right)>0$. Hence the first arrival time is not exponentially distributed with parameter $λ$, so $\left(Y_t,t≥0\right)$ not a $\mathrm{PP}(λ)$.
      Following the hint, we have for all $t≥0$ $$ 𝔼\left[Y_t\right]=𝔼\left[\sum_{n≥1}B_n\right]=\sum_{n≥1}ℙ\left(R_n≤t\right)=\sum_{n≥1}ℙ\left(T_n≤t\right)=λ t . $$ In particular, $ℙ\left(Y_k<∞\right)=1$ for all $k≥1$ and so $ℙ\left(Y_k<∞\right.$ for all $\left.k≥1\right)=1$. Since $t↦Y_t$ is (weakly) increasing, this already implies that on the same event $\left\{Y_k<∞\right.$ for all $\left.k≥1\right\}$, also $Y_t<∞$ for all $t≥0$
  1. State the Central Limit Theorem.
  2. Let $(R,S)$ be a pair of random variables with joint probability density function $$ f(r,s)=\begin{cases}\frac{1}{4}e^{-{|s|}},&(r,s)∈[-1,1]×ℝ\\0,&\text { otherwise. }\end{cases} $$ Also consider independent identically distributed random variables $\left(R_n,S_n\right),n⩾1$, with the same joint distribution as $(R,S)$.
    1. Find the marginal probability density functions of $R$ and $S$.
    2. For any $s∈ℝ$, determine $$ \lim _{n→∞}ℙ\left(\frac{1}{\sqrt{n\operatorname{var}(S)}}\sum_{k=1}^n S_n⩽s\right) $$
    3. For any $r,s∈ℝ$, show that $$ \lim _{n→∞}ℙ\left(\frac{1}{\sqrt{n\operatorname{var}(R)}}\sum_{k=1}^n R_n⩽r,\frac{1}{\sqrt{n\operatorname{var}(S)}}\sum_{k=1}^n S_n⩽s\right)=ℙ(W⩽r,Z⩽s) $$ for a pair of random variables $(W,Z)$ whose joint distribution you should determine.
  3. Consider the transformation $T:ℝ^2→ℝ^2$ given by $T(x,y)=(x-y,x+y)$. Let $(R,S)$ be as in (b) and $(X,Y)$ such that $(R,S)=T(X,Y)$.
    1. Derive the joint probability density function of $(X,Y)$.
    2. Find the marginal probability density functions of $X$ and $Y$.
    3. Find the correlation of $X$ and $Y$.
  1. If $V_n,n≥1$, are iid with mean $μ$ and variance $σ^2∈(0,∞)$, for all $v∈ℝ$$$ℙ\left(\frac{1}{\sqrt{n σ^2}}\sum_{k=1}^n\left(V_n-μ\right)≤v\right)→ℙ(Z≤v) \text{ as }n→∞$$where $Z∼N(0,1)$.
    1. Joint pdf factorises, so read off $f_R(r)=\frac{1}{2},r∈[-1,1],f_S(s)=\frac{1}{2}e^{-{|s|}},s∈ℝ$.
    2. Clearly $S_n,n≥1$, are iid with zero mean, by symmetry, and variance $\operatorname{var}(S)=𝔼\left[S^2\right]=2∫_0^∞s^2\frac{1}{2}e^{-s}d s=2∈(0,∞)$. Hence the CLT of (a) yields$$ℙ\left(\frac{1}{\sqrt{n\operatorname{var}(S)}}\sum_{k=1}^n S_n≤s\right)→ℙ(Z≤s),\text{ for all $s∈ℝ$, where $Z∼N(0,1)$.}$$
    3. $R_n,n≥1$, are also iid with zero mean and $\operatorname{var}(R)=\frac{1}{3}∈(0,∞)$. As $f$ factorises, $R$ and $S$, and hence their partial sums are independent and by CLT \begin{aligned} & ℙ\left(\frac{1}{\sqrt{n \operatorname{var}(R)}} \sum_{k=1}^n R_n≤r, \frac{1}{\sqrt{n \operatorname{var}(S)}} \sum_{k=1}^n S_n≤s\right) \\ &=ℙ\left(\frac{1}{\sqrt{n \operatorname{var}(R)}} \sum_{k=1}^n R_n≤r\right) ℙ\left(\frac{1}{\sqrt{n \operatorname{var}(S)}} \sum_{k=1}^n S_n≤s\right) \\ &→ℙ(W≤r) ℙ(Z≤s)=ℙ(W≤r, Z≤s) \end{aligned} where the joint distribution of $(W,Z)$ is determined by independence and marginal distributions $W,Z∼N(0,1)$
    1. Linear transformation $T$, bijective with Jacobian determinant $J(x, y)=1 × 1-(-1) × 1=2$. By the transformation formula for pdfs, $(X, Y)$ has joint pdf $$ f_{X, Y}(x, y)={|J(x, y)|} f_{R, S}(T(x, y))= \begin{cases}\frac{1}{2} e^{-{|x+y|}} & \text { if }{|x-y|} ≤ 1 \\ 0 & \text { otherwise. }\end{cases} $$
    2. By symmetry $X$ and $Y$ are identically distributed. Also, for ${|y|} ≥ \frac{1}{2}$, \begin{aligned} f_Y(y)=f_Y({|y|}) & =∫_{|y|-1}^{|y|+1} \frac{1}{2} e^{-x-{|y|}} d x=\frac{1}{2} e^{-{|y|}}\left(e^{-{|y|}+1}-e^{-{|y|}-1}\right) \\ & =\frac{1}{2}\left(e-\frac{1}{e}\right) e^{-2{|y|}} \end{aligned} and for ${|y|} ≤ \frac{1}{2}$, we split into two integrals over $({|y|}-1,-{|y|})$ and $(-{|y|},{|y|}+1)$ : \begin{aligned} f_Y(y)=f_Y({|y|}) & =\left(\frac{1}{2}-\frac{1}{2} e^{2{|y|}-1}\right)+\left(-\frac{1}{2} e^{-2{|y|}-1}+\frac{1}{2}\right) \\ & =1-\frac{1}{e} \cosh (2 y) . \end{aligned}
    3. $\operatorname{var}(Y)=\operatorname{var}(X)=\operatorname{var}\left(\frac{1}{2} S+\frac{1}{2} R\right)=\frac{1}{4} \operatorname{var}(S)+\frac{1}{4} \operatorname{var}(R)=\frac{7}{12}$ by independence. $\operatorname{cov}(X, Y)=\operatorname{cov}\left(\frac{1}{2}(S+R), \frac{1}{2}(S-R)\right)=\frac{1}{4}(\operatorname{var}(S)-\operatorname{var}(R))=\frac{5}{12}$, so $ρ=\frac{5}{7}$.
  1. Consider a Markov chain on a countable state space $S$ and let $i ∈ S$.
    1. Define the notions of recurrence and positive recurrence of $i$.
    2. Suppose that $i$ is positive recurrent. State, without proof, the ergodic theorem for the long-term proportion of time the Markov chain spends in state $i$.
  2. An urn contains a total of $N ⩾ 2$ balls, some white and the others black. Each step consists of two parts. A first ball is chosen at random and removed. A second ball is then chosen at random from those remaining. It is returned to the urn along with a further ball of the same colour. Denote by $Y_n$ the number of white balls after $n ⩾ 0$ steps.
    1. Explain briefly why $\left(Y_n, n ⩾ 0\right)$ is a Markov chain and determine its state space and transition matrix.
    2. Determine the communicating classes of this Markov chain and say whether their states are recurrent, and whether they are aperiodic. Justify your answers.
    3. Find all stationary distributions of this Markov chain.
  3. Now consider a Markov chain $\left(Z_n, n ⩾ 0\right)$, on $I=\{0,1,2, …, N\}$ with the transition matrix $P$ whose non-zero entries are $$ p_{k, j}= \begin{cases}\frac{N-k}{N} \frac{k+1}{N+1} & \text { if } j=k+1, \\ \frac{N-k}{N} \frac{N-k}{N+1}+\frac{k}{N} \frac{k}{N+1} & \text { if } j=k, \\ \frac{k}{N} \frac{N-k+1}{N+1} & \text { if } j=k-1 .\end{cases} $$
    1. Show that the uniform distribution is stationary for this Markov chain. Hence, or otherwise, determine all stationary distributions of this Markov chain.
    2. For a state $k ∈ I$, consider the successive visits $$ V_1^{(k)}=\inf \left\{n ⩾ 1: Z_n=k\right\}   \text { and }   V_{m+1}^{(k)}=\inf \left\{n ⩾ V_m^{(k)}+1: Z_n=k\right\},   m ⩾ 1 . $$ Explain why visits to $k$ occur in groups of independent geometrically distributed consecutive visits, and determine the parameter of this geometric distribution.
    3. Determine the expected time between two groups of visits to state $k$.
    4. Is the following statement true or false? "For any two states $k_1 ≠ k_2$, there is, on average, one visit to $k_2$ between the first and second visits to $k_1$. " Provide a proof or counterexample.
    1. Denote the MC by $\left(X_n\right)$. Let $H^{(i)}=\inf \left\{n ≥ 1: X_n=i\right\}$. Then $i$ is recurrent, resp. positive recurrent, if $ℙ_i\left(H^{(i)}<∞\right)=1$, resp. $m_i:=𝔼\left[H^{(i)}\right]<∞$.
    2. In the given setting, for all $j ∈ S$ with $ℙ_j\left(H^{(i)}<∞\right)=1$, we have $$ ℙ_j\left(n^{-1} \#\left\{1 ≤ k ≤ n: X_k=i\right\} → 1 / m_i\right)=1 . $$
  2. S: (i)-(ii) standard, (iii) new but similar to other examples.
    1. Only the numbers $Y_n$ of white and $N-Y_n$ of black balls in the urn at time $n$ are relevant for the future evolution of the urn, not how we got to this composition. The state space is $I=\{0,1,2, …, N\}$ and the non-zero entries of the transition matrix are $$ p_{k, j}= \begin{cases}\frac{N-k}{N} \frac{k}{N-1} & \text { if } j=k+1, \\ \frac{N-k}{N} \frac{N-k-1}{N-1}+\frac{k}{N} \frac{k-1}{N-1} & \text { if } j=k, \\ \frac{k}{N} \frac{N-k}{N-1} & \text { if } j=k-1 .\end{cases} $$
    2. $\{0\}$ and $\{N\}$ are recurrent aperiodic singleton classes as $p_{0,0}=p_{N, N}=1$, while $\{1, …, N-1\}$ is a transient class since $p_{i, i-1}=p_{i, i+1}>0$ for all $1 ≤ i ≤ N-1$. It is aperiodic since also $p_{i, i}>0$ for all $1 ≤ i ≤ N-1$.
    3. Stationary distributions vanish on transient classes. On the other hand, all distributions with this property, $(λ, 0,0, …, 0,1-λ), 0 ≤ λ ≤ 1$, are stationary.
  3. (i) is standard, (ii) is new, but an elementary observation, (iii) is similar to other questions where return times are split up. (iv) is new.
    1. Let $π_i=1 /(N+1), 0 ≤ i ≤ N$, be the uniform distribution on $I$. We need to check $π P=π$. Indeed, we have \begin{aligned} (π P)_j & =π_{j-1} p_{j-1, j}+π_j p_{j, j}+π_{j+1} p_{j+1, j} \\ & =\frac{1}{N+1}\left(\frac{N-(j-1)}{N} \frac{(j-1)+1}{N+1}+\frac{N-j}{N} \frac{N-j}{N+1}+\frac{j}{N} \frac{j}{N+1}+\frac{j+1}{N} \frac{N-(j+1)+1}{N+1}\right)=\frac{1}{N+1}=π_j, \end{aligned} for all $0 ≤ j ≤ N$, with the conventions $p_{-1,0}=p_{N+1, N}=0$ that are appropriately included above. $π$ is unique since this Markov chain is irreducible, by the same argument as in (b)(ii) here also including $p_{0,1}=p_{N, N-1}>0$.
    2. By the Markov property, the first step away from $k$ is like the first success in a sequence of Bernoulli trials with success probability $q_k:=1-p_{k, k}$.
    3. From (i) and lectures, the expected return time is $𝔼_k\left[V_1^{(k)}\right]=m_k=1 / π_k=N+1$. Denote by $A$ the event that the first step is away from $k$. We want to find $x=𝔼_k\left[V_1^{(k)} ∣ A\right]$. Clearly $𝔼_k\left[V_1^{(k)} ∣ A^c\right]=1$. By the Law of Total Probability, $$ N+1=m_k=𝔼_k\left[V_1^{(k)} ∣ A\right] ℙ(A)+𝔼_k\left[V_1^{(k)} ∣ A^c\right] ℙ\left(A^c\right)=x q_k+\left(1-q_k\right) $$ Hence $x=\left(N+q_k\right) / q_k=1+N / q_k$.
    4. This statement is true. Assume there are $k_1$ and $k_2$ with $r ≠ 1$ visits to $k_2$ between two consecutive visits to $k_1$, on average. By the strong Markov property, the numbers $W_k$ of visits are iid. Since $V_m^{\left(k_1\right)} → ∞$ a.s., the ergodic theorem says the asymptotic proportion of time in $k_2$ up to $V_m^{\left(k_1\right)}$ is $1 /(N+1)$. But SLLN says $$ \frac{W_1+⋯+W_m}{V_m^{\left(k_1\right)}}=\frac{W_1+⋯+W_m}{m} \frac{m}{V_m^{\left(k_1\right)}} → r \frac{1}{m_{k_1}}=\frac{r}{N+1}   \text { a.s., as } m → ∞ . $$ This contradicts the uniqueness of limits.