Past exam of the options I course of the University of Oxford / 2009 / M / M1

Pointwise,Taking expectations and dividing by $a^2>0$ gives the second-moment form of Markov's inequality:

The inverse transformation is $x=uv$, $y=u(1-v)$, with support $u>0$, $0<v<1$. Its absolute Jacobian determinant isThe original joint probability density function, by independence, is ${\lambda }^2{e}^{-\lambda (x+y)}$ on the positive quadrant. The change of variables formula therefore givesIntegrating over $v$ givesthe gamma distribution of shape two and rate $\lambda$. Integrating over $u$ gives $f_V(v)=1$ on $(0,1)$. The factorization also shows that $U$ and $V$ are independent.

The preceding marginal calculation makes each $V_i$ uniformly distributed on $(0,1)$, with expectation $1/2$ and variance $1/12$. By independence,Apply part (a) to the centred random variable $Z=S_n/n-1/2$ to obtainThis is Chebyshev's inequality in this setting.