Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / M / 1 (M1)

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.