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For a centered square-integrable martingale with $v=\mathbb{E}{M}_n^2$ and $\lambda >0$, apply the Doob maximal inequality for a nonnegative submartingale to $(M_k+c{)}^2$ for $c\ge 0$. Conditional Jensen inequality gives the submartingale property, and crossing $\lambda$ forces its square above $(\lambda +c{)}^2$. The bound $(v+c^2)/(\lambda +c{)}^2$ is minimized by $c=v/\lambda$. This extends the Cantelli inequality from one random variable to a martingale maximum, without requiring independent increments.

Put $v=\mathbb{E}{X}_n^2$. For $c\ge 0$, conditional Jensen inequality applied to the convex function $x\mapsto (x+c{)}^2$ shows thatis a nonnegative submartingale. Its integrability follows from the square integrability of $X_k$. If $X_k\ge \lambda$, then $Z_k\ge (\lambda +c{)}^2$, since $c\ge 0$. The Doob maximal inequality for a nonnegative submartingale yieldswhere zero mean removes the cross term. For completeness, the maximal inequality follows by stopping at the first crossing: on the event of a crossing at $k$, the submartingale property gives $\mathbb{E}[Z_n{1}_{\{T=k\}}]\ge \mathbb{E}[Z_k{1}_{\{T=k\}}]$. Sum over $k\le n$, and use nonnegativity on the event of no crossing.

The derivative of the last ratio isFor $v>0$, the minimum over $c\ge 0$ is attained at $c=v/\lambda$. Substitution gives the one-sided maximal inequality for a centered square-integrable martingale:If $v=0$, $X_n=0$ almost surely and $X_k=\mathbb{E}(X_n\mid {\mathcal{F}}_k)=0$ almost surely for every $k\le n$, so the bound also holds. The optimization is the same one underlying the Cantelli inequality, but the submartingale argument controls the entire finite-time maximum.

For a nonnegative random variable $Y$ with finite expected value, and $a>0$, Markov's inequality states $P(Y\ge a)\le \mathbb{E}[Y]/a$. Indeed $Y\ge a{1}_{\{Y\ge a\}}$, and taking expected values proves the bound. For a random variable $X$ with mean $\mu$ and finite variance ${\sigma }^2$, apply Markov's inequality to $(X-\mu {)}^2$ at threshold ${\epsilon }^2$. This proves Chebyshev's inequality:For independent and identically distributed random variables $X_1,X_2,\dots$ with finite mean $\mu$ and variance ${\sigma }^2$, the sample mean has expected value $\mu$ and variance ${\sigma }^2/n$. ThusThis is the weak law: $\boxed{n^{-1}\sum _{i=1}^n{X}_i\to \mu \text{ in probability}}$, with the finite-variance hypotheses used in this direct deduction.

The more general weak law of large numbers needs only $\mathbb{E}|X_1|<\infty$. To obtain that version, truncate $Y_i^{(K)}=X_i{1}_{\{|X_i|\le K\}}$. For fixed $K$ the bounded variables obey the result just proved. Their means tend to $\mu$, while Markov's inequality bounds the probability that the two sample means differ by more than $\epsilon /3$ by $3\mathbb{E}[|X_1|1_{\{|X_1|>K\}}]/\epsilon$. First take $n\to \infty$ and then $K\to \infty$; the integrable tail vanishes, proving the full integrable i.i.d. version as well.

Finally assume $\mathbb{E}[X]=0$ and $\mathrm{Var}(X)={\sigma }^2$. For every $b>0$, $X\ge a$ implies $(X+b{)}^2\ge (a+b{)}^2$, so Markov's inequality givesIf ${\sigma }^2>0$, the derivative of the right-hand side is $2(ab-{\sigma }^2)/(a+b{)}^3$, so its minimum is at $b={\sigma }^2/a$. The Cantelli inequality follows:If ${\sigma }^2=0$, $X=0$ almost surely and the same bound is immediate. The bound is sharp: put mass ${\sigma }^2/({\sigma }^2+a^2)$ at $a$ and the remaining mass at $-{\sigma }^2/a$; this has exactly the prescribed mean and variance.