Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / M / 3 (M3)

The definition of convergence in probability isThe variables are defined on a common probability space so that these difference events make sense.

Pointwise, $X\ge a{1}_{\{X\ge a\}}$, since $X$ is nonnegative. Taking expectations and dividing by $a>0$ proves the Markov inequalityIf the expectation is infinite, the inequality remains true in the extended-real sense.

Let ${\overline{X}}_n=n^{-1}{\sum }_i{X}_i$ and write ${\sigma }^2=\mathrm{Var}(X_1)<\infty$. Independence gives $\mathbb{E}{\overline{X}}_n=\mu$ and $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. Apply part (b) to the nonnegative variable $({\overline{X}}_n-\mu {)}^2$, which is the Chebyshev inequality:Thus ${\overline{X}}_n\to \mu$ in probability, proving the finite-variance weak law of large numbers.

Equality forces $\mathbb{E}X<\infty$ because its proposed value is at most $a$. The random variable $Y=X-a{1}_{\{X\ge a\}}$ is nonnegative and has expectation zero, so $Y=0$ almost surely. On $\{X<a\}$ it equals the strictly positive variable $X$, making that event null. On $\{X\ge a\}$ it equals $X-a$, forcing $X=a$ there. HenceThis is the equality in Markov's inequality under strict positivity. The reasoning even works without the discreteness assumption.