Sheet 3

Exercise 1 (Martingales Warm-up).

Do enough of these exercises to feel comfortable with them.

1. Assume that $X_n$ is an integrable $({ℱ}_n)$ adapted stochastic process which is decreasing, i.e.

   $X_{n+1}\le X_n\mathit{\hspace{1em}}\text{a.s.}.$

   Prove that $X_n$ is a supermartingale.

   Remark.

   Similarly an adapted increasing process is a submartingale.

   Solution.

   We simply use the monotonicity of the conditional expectation to obtain the last property of supermartingales

   $𝔼[X_{n+1}|{ℱ}_n]$ $\le 𝔼[X_n|{ℱ}_n]=X_n.\mathit{∎}$

2. Let ${(X_n)}_{n\in \mathbb{N}}$, ${(Y_n)}_{n\in \mathbb{N}}$ be ${ℱ}_n$ (sub-)martingales, prove that

   1. ${(X_n+Y_n)}_{n\in \mathbb{N}}$ is a (sub-)martingale.

      Solution.

      We have and ${ℱ}_n$-adapted follows from the fact that sums of measurable functions are measurable. What is left to show is

      $𝔼[X_{n+1}+Y_{n+1}|{ℱ}_n]=𝔼[X_{n+1}|{ℱ}_n]+𝔼[Y_{n+1}|{ℱ}_n]=X_n+Y_n$

      or the same with $\ge$ for submartingales respectively. ∎

   2. ${(X_n\vee Y_n)}_{n\in \mathbb{N}}$ is a submartingale,

      Solution.

      Adapted and integrable is easy to check, and the submartingale property follows from the monotonicity of conditional expectation

      $𝔼[X_{n+1}\vee Y_{n+1}|{ℱ}_n]$

      $=X_n\vee Y_n.\mathit{∎}$

3. Let ${(X_n)}_{n\in {\mathbb{N}}_0}$ be a $({ℱ}_n)$-submartingale. Let $T$ be $({ℱ}_n)$-stopping time.

   1. Suppose that ${(X_n)}_{n\in {\mathbb{N}}_0}$ is bounded and $T$ is a.s. finite (the meaning is different from ”$T$ is a.s. bounded”!). Prove that .

      Solution.

      Since a.s. we have

      $X_T=\underset{n\to \mathrm{\infty }}{lim}{X}_{T\wedge n}=\underset{n\to \mathrm{\infty }}{lim}{X}_n^T$

      where $X^T={(X_{T\wedge n})}_{n\in \mathbb{N}}$ is a bounded submartingale, because

      • as ${ℱ}_{n\wedge T}\subseteq {ℱ}_n$ as $n\wedge T\le n$ therefore $X^T$ is adapted

      • we have integrability due to

        $𝔼[|X_n^T|]$ $=𝔼\left[{\displaystyle \sum _{k=1}^n}|X_k|{\mathrm{𝟙}}_{T=k}\right]+𝔼\left[{\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}|X_n|{\mathrm{𝟙}}_{T=k}\right]$

        $\le 𝔼\left[{\displaystyle \sum _{k=1}^n}|X_k|\right]+𝔼[|X_n|{\mathrm{𝟙}}_{T>n}]$

      • Lastly we have the submartingale property

        $𝔼[X_{n+1}^T-X_n^T|{ℱ}_n]$ $=𝔼[(X_{n+1}^T-X_n^T)({\mathrm{𝟙}}_{T\le n}+{\mathrm{𝟙}}_{T>n})|{ℱ}_n]$

        $=𝔼[(X_{n+1}^T-X_n^T){\mathrm{𝟙}}_{T>n}|{ℱ}_n]$

        $={\mathrm{𝟙}}_{T>n}\underset{\ge 0}{\underbrace{𝔼[X_{n+1}^T-X_n^T|{ℱ}_n]}}$

      As $X_n$ is bounded, $X_T$ is bounded and therefore . By dominated convergence we further have

      $𝔼[X_T]$ $=𝔼[\underset{n\to \mathrm{\infty }}{lim}{X}_n^T]=\underset{n\to \mathrm{\infty }}{lim}𝔼[X_n^T]\ge 𝔼[X_0^T]=𝔼[X_0]\mathit{∎}$

   2. Suppose that there exists a constant $K>0$ such that for $\mathbb{P}$-a.e. $\omega \in \mathrm{\Omega }$,

      $$|X_n(\omega )-X_{n-1}(\omega )|\le K,\forall n\in \mathbb{N}.$$

      We also suppose that . Show that .

      Solution.

      $X^T$ is a submartingale for the same reason as before, and we can again use dominated convergence if we find an integrable upper bound. This upper bound is $TK+|X_0|$ because

      $X_n^T$ $=X_0+{\displaystyle \sum _{k=0}^{T\wedge n-1}}{X}_{k+1}^T-X_k^T$

      $\le |X_0|+{\displaystyle \sum _{k=0}^{T-1}}K=|X_0|+TK.\mathit{∎}$

   Hint.

   Compare this with statements about martingales in the lectures and use the bounded stopping time $T\mathrm{\wedge }M$ for some constant $M\mathrm{\in }\mathbb{N}$ and the dominated convergence theorem.

Exercise 2 (Conditional Expectation).

Recall that a gamma distribution with parameter $c>0$ and $\theta >0$ has density:

$$\frac{{\theta }^c}{\mathrm{\Gamma }(c)}{x}^{c-1}{e}^{-\theta x}{\mathrm{𝟙}}_{x>0}.$$

1. Let $X,Y$ be two independent exponential random variables with parameter $\theta >0$ and $Z=X+Y$. Determine the conditional distribution of $X$ given $Z=z$ (i.e. the Markov kernel $(A,z)\mapsto \mathbb{P}(X\in A|Z=z)$).

   Solution.

   For all $g$ non-negative and measurable functions we have that:

   $𝔼[h(X)g(Z)]$ $=𝔼[h(X)g(X+Y)]$

   $={\displaystyle {\int }_{{ℝ}^2}}h(x)g(x+y)f_X(x)f_Y(y)dxdy$

   $={\displaystyle {\int }_{{ℝ}^2}}h(x)g(x+y)\theta \exp (-\theta x){\mathrm{𝟙}}_{\{x\ge 0\}}\theta \exp (-\theta y){\mathrm{𝟙}}_{\{y\ge 0\}}dxdy$

   $={\displaystyle {\int }_{{ℝ}^2}}h(x)g(x+y){\theta }^2\exp (-\theta (x+y)){\mathrm{𝟙}}_{\{x\ge 0\}}{\mathrm{𝟙}}_{\{y\ge 0\}}dxdy$

   $\stackrel{z=x+y}{=}{\displaystyle {\int }_{{ℝ}^2}}h(x)g(z){\theta }^2\exp (-\theta z))\underbrace{{\mathrm{𝟙}}_{\{x\ge 0\}}{\mathrm{𝟙}}_{\{z-x\ge 0\}}}{}_{={\mathrm{𝟙}}_{z\ge 0}{\mathrm{𝟙}}_{z\ge x\ge 0}}\mathrm{d}x\mathrm{d}y$

   $={\displaystyle {\int }_{ℝ}}g(z){\theta }^2{e}^{-\theta z}{\mathrm{𝟙}}_{\{z\ge 0\}}z\left({\displaystyle \frac{1}{z}}{\displaystyle {\int }_0^z}h(x)dx\right)dz$

   $=𝔼\left[g(Z)\left({\displaystyle \frac{1}{Z}}{\displaystyle {\int }_0^Z}h(u)du\right)\right]$

   Since the last term is ${ℱ}_Z$-measurable we can conclude that $𝔼[h(X)|Z]={\int }_0^Z\frac{1}{Z}h(u)du$. Using $\mathbb{P}(X\in A|Z)=𝔼[{\mathrm{𝟙}}_A(X)|Z]$ means that for $A\in ℬ(ℝ)$, we get

   $\mathbb{P}(X\in A|Z=z)={\displaystyle \int }{\mathrm{𝟙}}_A{\displaystyle \frac{1}{z}}{\mathrm{𝟙}}_{[0,z]}\mathrm{d}u={\displaystyle \frac{\lambda ([0,z]\cap A)}{z}}.$

   Hence, the conditional distribution of $X$ given $Z=z$ is $𝒰[0,z]$. ∎

2. Conversely, let $Z$ be a random variable with gamma distribution with parameter $(2,\theta )$, and suppose $X$ is a random variable whose conditional distribution given $Z=z$ is uniform on $[0,z]$, for $z>0$. Prove that $X$ and $Z-X$ are independent with exponential distribution $\text{Exp}(\theta )$.

   Solution.

   With kernels we can easily write

   $𝔼[h(X,Z)|Z]$ $\stackrel{\text{Thm. 5.3.5}}{=}{\displaystyle {\int }_{ℝ}}h(u,Z)\mathbb{P}(X\in du|Z)$ (1)

   $={\displaystyle {\int }_{ℝ}}h(u,Z){\displaystyle \frac{1}{Z}}{\mathrm{𝟙}}_{\{0\le u\le Z\}}du$ (2)

   where $(A,z)\mapsto \mathbb{P}(X\in A|z)$ is the regular conditional distribution defined from the unique kernel we get via the disintegration theorem. Essentially doing all the same calculations backwards we get

   $𝔼[f(X)g(Z-X)]$ $=𝔼\left[𝔼[f(X)g(Z-X)|Z]\right]$

   $=𝔼\left[{\displaystyle {\int }_{ℝ}}f(u)g(Z-u){\displaystyle \frac{1}{Z}}{\mathrm{𝟙}}_{\{0\le u\le Z\}}du\right]$

   $={\displaystyle {\int }_{{ℝ}^2}}f(u)g(z-u){\displaystyle \frac{1}{z}}{\mathrm{𝟙}}_{\{0\le u\le z\}}\theta z{e}^{-\theta z}{\mathrm{𝟙}}_{z\ge 0}dudz$

   $={\displaystyle {\int }_{{ℝ}^2}}f(u)g(y){\theta }^2{e}^{-\theta (u+y)}{\mathrm{𝟙}}_{y\ge 0}{\mathrm{𝟙}}_{u\ge 0}dudy$

   $={\displaystyle {\int }_{ℝ}}f(u)\theta e^{-\theta u}{\mathrm{𝟙}}_{u\ge 0}du{\displaystyle {\int }_{ℝ}}g(y)\theta e^{-\theta y}{\mathrm{𝟙}}_{y\ge 0}dy.$

   This shows that $(X,Z-X)$ are independent $\text{Exp}(\theta )$ random variables. In more detail: Use $g\equiv 1$ to show $X\sim \text{Exp}(\theta )$. And similarly $f\equiv 1$ to show $Z-X\sim \text{Exp}(\theta )$. Independence follows using indicator functions in place of $g,f$. ∎

Exercise 3 (Branching Process Tools).

Assume $X={(X_n)}_n$ are identically distributed, independent and integrable random variables. And assume $N$ is an integrable ${\mathbb{N}}_0$-valued random variable independent of $X$ and $S_n:={\sum }_{k=1}^n{X}_i$.

1. (Wald’s equation) Prove that

   $𝔼[S_N]=𝔼[N]𝔼[X_1]$

   Hint.

   Use indicators and Fubini.

   Solution.

   The claim follows from Fubini and independence:

   $𝔼\left[{\displaystyle \sum _{k=1}^N}{X}_i\right]$ $=𝔼\left[{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\mathrm{𝟙}}_{N=n}{\displaystyle \sum _{k=1}^n}{X}_i\right]={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}𝔼\left[{\mathrm{𝟙}}_{N=n}{\displaystyle \sum _{k=1}^n}{X}_i\right]$

   $\stackrel{\text{indep.}}{=}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}𝔼[{\mathrm{𝟙}}_{N=n}]\underset{=n𝔼[X_1]}{\underbrace{𝔼\left[{\displaystyle \sum _{k=1}^n}{X}_i\right]}}$

   $=𝔼[X_1]{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}𝔼[n{\mathrm{𝟙}}_{N=n}]$

   $=𝔼[X_1]𝔼\left[N{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\mathrm{𝟙}}_{N=n}\right]$

   $=𝔼[X_1]𝔼[N]\mathit{∎}$

2. (Blackwell-Girshick) Further assume that $X_n,N\in {ℒ}^2$. Prove that

   1. $𝔼[S_N^2]=𝔼[N]\text{Var}[X_1]+𝔼[N^2]𝔼{[X_1]}^2$

   2. $\text{Var}[S_N]=𝔼[N]\text{Var}[X_1]+\text{Var}[N]𝔼{[X_1]}^2$

   Hint.

   Same trick as with Wald’s equation.

   Solution.

   Similarly to Wald’s equation we have

   $𝔼[S_N^2]$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}𝔼[{\mathrm{𝟙}}_{N=n}{S}_n^2]$

   $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}𝔼[{\mathrm{𝟙}}_{N=n}]𝔼[S_n^2]$

   $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}𝔼[{\mathrm{𝟙}}_{N=n}](\underset{=n\text{Var}[X_1]}{\underbrace{\text{Var}[S_n]}}+{(\underset{=n𝔼[X_1]}{\underbrace{𝔼[S_n]}})}^2)$

   $=𝔼\left[{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\mathrm{𝟙}}_{N=n}n\right]\text{Var}[X_1]+𝔼\left[{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\mathrm{𝟙}}_{N=n}{n}^2\right]𝔼{[X_1]}^2$

   $=𝔼[N]\text{Var}[X_1]+𝔼[N^2]𝔼{[X_1]}^2$

   which directly implies (ii)(a). (ii)(b) follows using Wald’s equation as

   $\text{Var}[S_N]$ $=𝔼[S_N^2]-𝔼{[S_N]}^2$

   $=𝔼[N]\text{Var}[X_1]+𝔼[N^2]𝔼{[X_1]}^2-{(𝔼[N]𝔼[X_1])}^2$

   $=𝔼[N]\text{Var}[X_1]+\text{Var}[N]𝔼{[X_1]}^2\mathit{∎}$

Exercise 4 (Intro to Harmonic Functions).

Let $S_n≔{\sum }_{i=1}^n{Y}_i$ be a symmetric random walk, i.e. the jump sizes $(Y_n,n\ge 1)$ are a sequence of i.i.d. random variables with $\mathbb{P}(Y_n=1)=1/2$ and $\mathbb{P}(Y_n=-1)=1/2$. Let ${𝒢}_n:=\sigma (Y_1,Y_2,\mathrm{\dots }{Y}_n)$ with ${𝒢}_0:=\{\mathrm{\varnothing },\mathrm{\Omega }\}$.

1. Let $\lambda \in ℝ$. Find a constant $c\in ℝ$ such that $\exp {(\lambda S_n-cn)}_{n\in \mathbb{N}}$ is a $({𝒢}_n)$-martingale.

   Solution.

   To get the martingale property

   $𝔼[\exp (\lambda S_{n+1}-c(n+1))|{𝒢}_n]$ $=𝔼[\exp (\lambda S_n-cn)\exp (-c)\exp (\lambda Y_{n+1})|{𝒢}_n]$

   $=\exp (\lambda S_n-cn)\exp (-c)𝔼[\exp (\lambda Y_{n+1})|{𝒢}_n]$

   $=\exp (\lambda S_n-cn)\exp (-c)𝔼[\exp (\lambda Y_0)]$

   we need $\exp (c)=𝔼[\exp (\lambda Y_0)]$, i.e.

   $c$ $=\log (𝔼[\exp (\lambda Y_0)])=\log (\frac{1}{2}(\exp (\lambda )+\exp (-\lambda ))).\mathit{∎}$

2. Prove that ${(S_n^2-n)}_{n\in {\mathbb{N}}_0}$ is a $({𝒢}_n)$-martingale.

   Solution.

   We simply check the three properties of a martingale

   • ${(S_n^2-n)}_{n\in \mathbb{N}}$ is by definition of ${𝒢}_n$ adapted.

   • $𝔼[\left|S_n^2-n\right|]\le 𝔼[\left|S_n^2\right|+n]\le 𝔼[{\left|S_n\right|}^2]+n\le {\left|n\right|}^2+n\le \mathrm{\infty }$

   • Lastly we check the martingale property

     $𝔼\left[S_{n+1}^2-(n+1)|{𝒢}_n\right]$

     $=𝔼\left[\left({\displaystyle \sum _{i=1}^{n+1}}{Y}_i\right)\right|{𝒢}_n]-n-1$

     $=𝔼[{\left({\displaystyle \sum _{i=1}^n}{Y}_i\right)}^2+2Y_{n+1}{\displaystyle \sum _{i=1}^n}{Y}_i+{(Y_{n+1})}^2|{𝒢}_n]-n-1$

     $={\left({\displaystyle \sum _{i=1}^n}{Y}_i\right)}^2+2𝔼[Y_{n+1}|{𝒢}_n]{\displaystyle \sum _{i=1}^n}{Y}_i+𝔼[{(Y_{n+1})}^2|{𝒢}_n]-n-1$

     $={\left({\displaystyle \sum _{i=1}^n}{Y}_i\right)}^2+2\underset{=0}{\underbrace{𝔼[Y_{n+1}]}}{\displaystyle \sum _{i=1}^n}{Y}_i+\underset{=1}{\underbrace{𝔼[{(Y_{n+1})}^2]}}-1-n$

     $=S_n^2-n\mathit{∎}$

3. Prove that ${(S_n^3-3n{S}_n)}_{{\mathbb{N}}_0}$ is a $({𝒢}_n)$-martingale.

   Solution.

   We again check the definition

   • Adaptedness follows again from def of ${𝒢}_n$.

   • $𝔼[\left|S_n^3-3n{S}_n\right|]\le 𝔼[|S_n^3|]+3n𝔼[|S_n|]\le n^3+3n^2$

   • First, observe that

     $S_{n+1}^3={\left({\displaystyle \sum _{i=1}^{n+1}}{Y}_i\right)}^3={\left({\displaystyle \sum _{i=1}^n}{Y}_i\right)}^3+3{\left({\displaystyle \sum _{i=1}^n}{Y}_i\right)}^2{Y}_{n+1}+3\left({\displaystyle \sum _{i=1}^n}{Y}_i\right)Y_{n+1}^2+Y_{n+1}^3.$

     Using this we get

     $𝔼[S_{n+1}^3-3(n+1)S_{n+1}|{𝒢}_n]$

     $=𝔼[S_{n+1}^3|{𝒢}_n]-3(n+1)𝔼[S_{n+1}|{𝒢}_n]$

     $=S_n^3-3n{S}_n.\mathit{∎}$

4. Find a polynomial $P(s,n)$ with degree $4$ on $s$ and degree $2$ on $n$, such that ${(P(S_n,n))}_{n\in {\mathbb{N}}_0}$ is a $({𝒢}_n)$-martingale.

   Solution.

   We are going to use (v) to guess the correct polynomial. Before we start consider the simple polynomial $s^k$. We have

   ${(s+1)}^k+{(s-1)}^k={\displaystyle \sum _{i=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)s^{k-i}(1+{(-1)}^i)=2{\displaystyle \sum _{i=0}^{\lfloor \frac{k}{2}\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)s^{k-2i}$

   So when striving for $P(s+1,n+1)+P(s-1,n+1)=P(s,n)$ we only need to consider even (odd) exponents, if the degree of the polynomial is even (or odd respectively) (cf. (ii) (iii)). So let us start guessing with $P(s,n)=s^4+a(n)s^2+c(n)$ where $a$ and $c$ are polynomials with a degree of at most $2$. From this we obtain

   $P(s+1,n+1)+P(s-1,n+1)$

   $=2[s^4+\underset{\stackrel{!}{=}a(n)}{\underbrace{(6+a(n+1))}}{s}^2+\underset{\stackrel{!}{=}c(n)}{\underbrace{(1+a(n+1))+c(n+1)}}].$

   To solve this systematically we assume $a(n)=a_2{n}^2+a_{1}n+a_0$, which implies

   $6+a_2{(n+1)}^2+a_1(n+1)+a_0\stackrel{!}{=}{a}_2{n}^2+a_{1}n+a_0$

   $\iff a_2({(n+1)}^2-n^2)+(a_1+6)\stackrel{!}{=}0$

   $\iff 2a_{2}n+(a_0+a_1+6)\stackrel{!}{=}0$

   this is solved by $a_2=0$ and $a_1=-6$. So we have $a(n)=-6n$. With similar assumptions on $c$ we get

   $1-6(n+1)+[c_2{(n+1)}^2+c_1(n+1)+c_0]\stackrel{!}{=}{c}_2{n}^2+c_{1}n+c_0$

   $\iff c_2((n^2+2n+1)-n^2)+-6n+[c_1-5]\stackrel{!}{=}0$

   $\iff (2c_2-6)n+[c_2+c_1-5]\stackrel{!}{=}0.$

   This implies $c_2=3$ and therefore $c_1=2$. In total we have $P(s,n)=s^4-6n{s}^2+3n^2+2n$ fulfills (v). Therefore, $(S_n^4-6n{S}_n^2+3n^2+2n)$ is a martingale by (v). ∎

5. Prove that, in general, for a polynomial $P(x,y)$, the process ${(P(S_n,n))}_{n\in {\mathbb{N}}_0}$ is a $({𝒢}_n)$-martingale, if

   $$P(s+1,n+1)+P(s-1,n+1)=2P(s,n).$$

   Solution.

   • Adaptability follows from measurability of polynomials,

   • integrability from the triangle inequality and the fact that ${|S_n|}^k\le n^k$.

   • Lastly, note that $S_n$ is ${ℱ}_n$-measurable and $Y_{n+1}$ is independent of ${ℱ}_n$, this results in the martingale property

     $𝔼[P(S_{n+1},n+1)|{ℱ}_n]$ $=𝔼[P(S_n+Y_{n+1},n+1)|{ℱ}_n]$

     $=\frac{1}{2}P(S_n+1,n+1)+\frac{1}{2}P(S_n-1,n+1)$

     $=P(S_n,n)$

   We could have solved (ii) this way, as for $P(s,n)=s^2-n$, we have

   $P(s+1,n+1)+P(s-1,n+1)$ $={(s+1)}^2-(n+1)+{(s-1)}^2-(n+1)$

   $=2s^2+2-2n-2=2s^2-2n$

   $=2P(s,n).\mathit{∎}$

Exercise 5 (Poisson Process is Poisson).

In this exercise we are going to assume that the Poisson process ${(X_t)}_{t\ge 0}$ is defined as the number of events up to time $t$, where the waiting time between events is always exponentially distributed, i.e.

$X_t:=\max \left\{k\in {\mathbb{N}}_0:T_k:={\displaystyle \sum _{i=1}^k}{Y}_i\le t\right\},Y_i\stackrel{\text{iid}}{\sim }\text{Exp}(\lambda )$

Show that $X_t$ is Poisson distributed for every $t\ge 0$.

Hint.

The exponential distribution is a special case of the Gamma distribution, i.e. $\text{𝐸𝑥𝑝}\mathit{}\mathrm{(}\lambda \mathrm{)}\mathrm{=}\mathrm{\Gamma }\mathit{}\mathrm{(}\mathrm{1}\mathrm{,}\frac{\mathrm{1}}{\lambda }\mathrm{)}$. And since the Gamma distribution is stable under summation of iid Gamma distributed random variables, we get $T_k\mathrm{\sim }\mathrm{\Gamma }\mathit{}\mathrm{(}k\mathrm{,}\frac{\mathrm{1}}{\lambda }\mathrm{)}$. Using this fact you can either calculate $\mathbb{P}\mathrm{(}{X}_t\mathrm{\le }k\mathrm{)}$ using the cumulative distribution function of the Gamma distribution (which you might know or need to calculate), or you could calculate $\mathbb{P}\mathrm{(}{X}_t\mathrm{=}k\mathrm{)}$ by conditioning on $T_k$ in this sense

$\mathbb{P}(A)=𝔼[{\mathrm{𝟙}}_A]=𝔼[𝔼[{\mathrm{𝟙}}_A|T_k]].$

This avoids the usage of the cdf of the Gamma distribution.

Solution.

1. Using the fact that $T_{k+1}\sim \mathrm{\Gamma }(k+1,\frac{1}{\lambda })$ we get

   $\mathbb{P}(X_t\le k)$ $=\mathbb{P}(\text{no more than k events before t})$

   $=\mathbb{P}(\text{the (k+1)-event happens after t})$

   $=\mathbb{P}(T_{k+1}>t)$

   $=1-\mathbb{P}(T_{k+1}\le t)$

   $=1-{\displaystyle {\int }_0^t}{\displaystyle \frac{{\lambda }^{k+1}}{\mathrm{\Gamma }(k+1)}}{x}^k\exp (-\lambda x)𝑑x$

   $\stackrel{z=\lambda x}{=}1-{\displaystyle {\int }_0^{\lambda t}}{\displaystyle \frac{{\lambda }^{k+1}}{\mathrm{\Gamma }(k+1)}}{\left({\displaystyle \frac{z}{\lambda }}\right)}^k\exp (-z){\displaystyle \frac{1}{\lambda }}𝑑z$

   $=1-{\displaystyle {\int }_0^{\lambda t}}{\displaystyle \frac{1}{\mathrm{\Gamma }(k+1)}}{z}^k\exp (-z)𝑑z$

   By induction one can show that

   ${\displaystyle {\int }_0^t}{\displaystyle \frac{z^{k-1}}{\mathrm{\Gamma }(k)}}\exp (-z)𝑑z=1-\exp (-t){\displaystyle \sum _{n=0}^{k-1}}{\displaystyle \frac{t^n}{n!}}$

   where the induction step is simply partial integration:

   ${\displaystyle {\int }_0^t}{\displaystyle \frac{z^k}{\mathrm{\Gamma }(k+1)}}\exp (-z)𝑑z$ $={{\displaystyle \frac{z^k}{\mathrm{\Gamma }(k+1)}}(-\exp (-z))|}_0^t+{\displaystyle {\int }_0^t}{\displaystyle \frac{k{z}^{k-1}}{\mathrm{\Gamma }(k+1)}}\exp (-z)𝑑z$

   $=-\exp (-t){\displaystyle \frac{t^k}{\mathrm{\Gamma }(k+1)}}+{\displaystyle {\int }_0^t}{\displaystyle \frac{z^{k-1}}{\mathrm{\Gamma }(k)}}\exp (-z)𝑑z$

   $\stackrel{\text{ind.}}{=}-\exp (-t){\displaystyle \frac{t^k}{k!}}+1-\exp (-t){\displaystyle \sum _{n=0}^{k-1}}{\displaystyle \frac{t^n}{n!}}$

   $=1-\exp (-t){\displaystyle \sum _{n=0}^k}{\displaystyle \frac{t^n}{n!}}.$

   Therefore we have

   $\mathbb{P}(X_t\le k)=\exp (-\lambda t){\displaystyle \sum _{n=0}^k}{\displaystyle \frac{{(\lambda t)}^n}{n!}}$

   which implies that $X_t\sim \text{Poi}(\lambda )$.

2. Conditioning on $T_k$ we can calculate

   $\mathbb{P}(X_t=k)$

   $=𝔼[\exp \left(-\lambda (t-T_k)\right){\mathrm{𝟙}}_{T_k\le t}]$

   $=\exp (-\lambda t)𝔼[\exp (\lambda T_k){\mathrm{𝟙}}_{T_k\le t}]$

   Using $T_k\sim \mathrm{\Gamma }(k,\frac{1}{\lambda })$ we therefore get

   $\mathbb{P}(X_t=k)$ $=\exp (-\lambda t){\displaystyle {\int }_0^t}\exp (\lambda x){\displaystyle \frac{{\lambda }^k}{\mathrm{\Gamma }(k)}}{x}^{k-1}\exp (-\lambda x)𝑑x$

   $=\exp (-\lambda t){\displaystyle \frac{{\lambda }^k}{\mathrm{\Gamma }(k)}}\underset{=\frac{t^k}{k}}{\underbrace{{\displaystyle {\int }_0^t}{x}^{k-1}𝑑x}}$

   $=\exp (-\lambda t){\displaystyle \frac{{(\lambda t)}^k}{k!}}$

   which implies $X_t\sim \text{Poi}(\lambda )$.

∎