= Solution
<Convergence in probability> means that for every $\varepsilon>0$, $\mathbb P(|X_n-X|\geq\varepsilon)\to0$. The use of $>$ instead of $\geq$ gives the equivalent definition by replacing $\varepsilon$ with a smaller positive number.
For $Y\geq0$, the pointwise inequality $Y\geq a\mathbf1_{\{Y\geq a\}}$ gives, on taking <expected values>, <Markov's inequality>
$$
\mathbb P(Y\geq a)\leq\frac{\mathbb EY}{a}.
$$
This also holds if the expectation is infinite, though the resulting bound is then uninformative. Apply it to $Y=(X-\mu)^2$ with threshold $\varepsilon^2$ to obtain <Chebyshev's inequality>:
$$
\mathbb P(|X-\mu|\geq\varepsilon)\leq\sigma^2/\varepsilon^2.
$$
Set $S_n=n^{-1}\sum_{i=1}^n(Z_i-\mathbb EZ_i)$. Its mean is zero. <Independence> eliminates cross covariances, and boundedness gives $\operatorname{Var}(Z_i)\leq\mathbb EZ_i^2\leq1$. Hence
$$
\operatorname{Var}(S_n)=\frac1{n^2}\sum_{i=1}^n\operatorname{Var}(Z_i)\leq\frac1n,
\qquad\mathbb P(|S_n|\geq\varepsilon)\leq\frac1{n\varepsilon^2}\to0.
$$
Thus $\boxed{S_n\to0\text{ in probability}}$. Identical distributions were never needed.
Back to article page
