Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / M (Probability)

Pointwise $Z^2\ge a^2{1}_{\{|Z|\ge a\}}$. Taking expectations and dividing by $a^2>0$ proves the Markov inequality

The joint density of $X,Y$ is ${\lambda }^2{e}^{-\lambda (x+y)}$ for positive $x,y$. Inverting the transformation gives $x=uv$, $y=u(1-v)$, with absolute Jacobian determinant $u$ and domain $u>0$, $0<v<1$. The change of variables for probability densities givesThe joint density factors into the gamma density of $U$ and the uniform density of $V$ on $(0,1)$, so these two transformed variables are independent.

Part (b) gives $V_i\sim U(0,1)$, so $\mathbb{E}{V}_i=1/2$ and $\mathrm{Var}(V_i)=1/12$. Independence yields $\mathbb{E}(S_n/n)=1/2$ and $\mathrm{Var}(S_n/n)=1/(12n)$. Apply part (a) to $Z=S_n/n-1/2$:This is a direct Chebyshev inequality for the sample mean.

For $n=0$, the identity matrix gives the assertion. Suppose it holds at $n$. Conditioning on the state at time $n$ and using the Markov property givesThe nonnegative countable sum is justified by the law of total probability. Induction proves the $n$-step transition matrix is $P^n$.

Let $p_n=\mathbb{P}(X_n=1\mid X_0=1)$. Conditioning on the previous state gives $p_{n+1}=(1-a)p_n+b(1-p_n)=b+(1-a-b)p_n$, with $p_0=1$. Its fixed point is $b/(a+b)$; subtracting it leaves a geometric recursion. Therefore

After the first step, a chain starting at one enters the closed class $\{2,3\}$ with probability $p$, starting at two, or enters $\{4,5\}$ with probability $1-p$ and can never reach three. Within the first class, the probability of going from its first state to its second in $n-1$ steps is the complement of the formula in part (b). ThusAt $n=1$ this correctly gives zero.

The total eventual mass of the closed class $\{4,5\}$ is ${\lambda }_4+{\lambda }_5+(1-p){\lambda }_1$. The other closed class cannot contribute to state five. Since $|1-a-b|<1$, the two-state chain within the last class converges to its stationary distribution, with state-five mass $a/(a+b)$. Therefore

The iid central limit theorem states that if $Y_1,Y_2,\dots$ are independent and identically distributed with finite mean $\mu$ and variance $0<{\sigma }^2<\infty$, thenEquivalently, its cumulative distribution functions converge at every real argument to the standard normal cumulative distribution function.

The joint density of $X,Y$ is $(2\pi {)}^{-1}{e}^{-(x^2+y^2)/2}$. The polar-coordinate Jacobian is $r$, soIntegration over the angular coordinate gives $f_R(r)=r{e}^{-r^2/2}$ for $r\ge 0$ and zero otherwise. Thus $R$ has the Rayleigh distribution, while $\Theta$ is uniform and independent of $R$.

For $w\ge 0$, $\mathbb{P}(W\le w)=\mathbb{P}(R\le \sqrt{w})=1-e^{-w/2}$. Differentiation givesThis is an exponential distribution of rate $1/2$, with mean two and variance four.

The independent variables $W_i$ have mean two and variance four by part (b). Their standardization isThe central limit theorem therefore proves the claimed convergence to the standard normal law, whose continuous cumulative distribution function is $(2\pi {)}^{-1/2}{\int }_{-\infty }^x{e}^{-u^2/2}du$ at every real $x$.