Solution. Since \(\Var \left (\overline {X}\right ) = \sigma ^{2}/n\), Chebyshev gives \[P\left (\left |\overline {X}-\mu \right |\geq \varepsilon \right ) \leq \frac {\sigma ^{2}}{n\varepsilon ^{2}} \longrightarrow 0 ,\] which is the weak law. Setting \(\varepsilon = 0.1\sigma \), \[\frac {\sigma ^{2}}{n(0.1\sigma )^{2}} = \frac {100}{n} \leq 0.05 \quad \Longrightarrow \quad n \geq 2000 .\]

Chebyshev is deliberately crude. The central limit theorem gives \(1.96\sigma /\sqrt {n} = 0.1\sigma \), that is \(n\approx 385\) — about a fifth as many. The bound holds for every distribution with finite variance, which is what it is paying for.

Solution. Take \(X\) continuous with density \(f\), which is zero on \(x<0\). Then \begin {align*} E(X) &= \int _0^{\infty }xf(x)\,dx = \int _0^{a}xf(x)\,dx + \int _a^{\infty }xf(x)\,dx\\ &\geq \int _a^{\infty }xf(x)\,dx\\ &\geq \int _a^{\infty }af(x)\,dx = aP(X\geq a). \end {align*}

The first inequality drops a nonnegative integral; the second uses \(x\geq a\) on the range of integration. Dividing by \(a>0\), \[P(X\geq a)\leq \frac {E(X)}{a}.\] The discrete case is the same argument with sums in place of integrals; the indicator proof given for Markov’s inequality in the text covers both at once.

Solution. Since \((X-\mu )^{2}\geq 0\), Markov’s inequality applies to it with \(a=k^{2}\): \[P\left ((X-\mu )^{2}\geq k^{2}\right )\leq \frac {E\left [(X-\mu )^{2}\right ]}{k^{2}}.\] Since \((X-\mu )^{2}\geq k^{2}\) if and only if \(|X-\mu |\geq k\), \[P\left (|X-\mu |\geq k\right )\leq \frac {E\left [(X-\mu )^{2}\right ]}{k^{2}} = \frac {\sigma ^{2}}{k^{2}}.\]

Solution. Write \(\sigma ^{2}\) for the common variance, taken to be finite. Then \[\overline {X}_n = \frac {X_1+X_2+\cdots +X_n}{n},\qquad \Var \left (\overline {X}_n\right ) = \frac {\sigma ^{2}}{n}.\] By Chebyshev’s inequality, for any \(\varepsilon >0\), \[P\left (\left |\overline {X}_n-\mu \right |\geq \varepsilon \right ) \leq \frac {E\left [\left (\overline {X}_n-\mu \right )^{2}\right ]}{\varepsilon ^{2}} = \frac {\Var \left (\overline {X}_n\right )}{\varepsilon ^{2}} = \frac {\sigma ^{2}}{n\varepsilon ^{2}}.\] Therefore \[\lim _{n\rightarrow \infty }P\left (\left |\overline {X}_n-\mu \right |\geq \varepsilon \right ) \leq \lim _{n\rightarrow \infty }\frac {\sigma ^{2}}{n\varepsilon ^{2}} = 0 ,\] that is, \(\overline {X}_n\underset {P}{\longrightarrow }\mu \).

The question assumes only a finite mean. The conclusion still holds then — that is Khinchin’s form of the weak law — but its proof needs a truncation argument. A finite variance is what lets Chebyshev do all the work here.

Solution. Since \(E(X_k)=p\) and \(\Var (X_k)=p(1-p)\), \[E(Y_n) = \frac 1n\sum _{k=1}^{n}E(X_k) = \frac {np}{n} = p\] and \[\Var (Y_n) = \frac {1}{n^{2}}\sum _{k=1}^{n}\Var (X_k) = \frac {1}{n^{2}}\,np(1-p) = \frac {p(1-p)}{n}.\] By Chebyshev’s inequality, for any \(\varepsilon >0\), \[P\left (|Y_n-p|\geq \varepsilon \right )\leq \frac {\Var (Y_n)}{\varepsilon ^{2}} = \frac {p(1-p)}{n\varepsilon ^{2}}\longrightarrow 0\qquad (n\rightarrow \infty ).\] Therefore \(Y_n\underset {P}{\longrightarrow }p\): the proportion of successes converges to the probability of success.

As set, the question wrote \(Y_n=\sum _{k=1}^{n}X_k\) with limit \(\mu \). The sum itself has mean \(np\) and does not converge; it is the average that does, and its limit is the common mean \(p\).

Solution. Let \(S=\sum _{i=1}^{20}X_i\). Each \(X_i\) has \[E(X_i) = \frac {0+1}{2} = \frac 12,\qquad \Var (X_i) = \frac {(1-0)^{2}}{12} = \frac {1}{12},\] so \(E(S) = 20\times \frac 12 = 10\) and \(\Var (S) = 20\times \frac {1}{12} = \frac 53\).

(a)

By the central limit theorem, \[P(S\leq 12) = P\left (\frac {S-E(S)}{\sigma _S}\leq \frac {12-10}{\sqrt {5/3}}\right ) \approx P(Z\leq 1.549)\approx 0.9393 .\]

(b)

We want the \(x\) with \(P(S\leq x) = 0.90\): \[P\left (Z\leq \frac {x-10}{\sqrt {5/3}}\right ) = 0.90 \quad \Longrightarrow \quad \frac {x-10}{\sqrt {5/3}} = \Phi ^{-1}(0.90) = 1.2816 .\] Hence \[x = 10+1.2816\sqrt {5/3}\approx 11.65 .\]