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The three preceding probabilities satisfyTheir expected value and second moment areSubtracting the squared mean givesThe value $\mu =2(2/10)$ also follows from linearity of expectation. The variance includes the finite-population reduction caused by sampling without replacement.

Let $I_j$ be the indicator random variable of a black ball at draw $j$. The black count after three draws is $N(B_3)=I_1+I_2+I_3$. By linearity of expectation,Independence is not required for this step. This is also the mean of the corresponding hypergeometric distribution.

The $i$th screen distance is $x_0/2^i$, starting at $i=1$. The previous expected value therefore scales to $E[Y_i]=x_0\ln 3/(\pi 2^i)$. To justify the infinite linearity of expectation, note that $|\tan {\theta }_i|\le \sqrt{3}$ throughout the restricted angular interval. ThusThis deterministic integrable bound justifies the expectation of an absolutely summable random series, even without assuming independent firing angles. Summing the geometric series gives

Let $I_j$ be the indicator random variable of an apple in lunch $j$. The number remaining is $R=5-I_1-I_2-I_3$. Each expected value $E[I_j]=P(A_j)=1/2$. The linearity of expectation therefore givesAlthough the indicator random variables are dependent under sampling without replacement, the linearity of expectation needs no independence assumption.

Assume the three standard dice are fair and independent. For one die $X$, a discrete uniform distribution on $1,\dots ,6$ givesThe variance is $\mathrm{Var}X=\mathbb{E}{X}^2-(\mathbb{E}X{)}^2=35/12$. For the sum $Q=X_1+X_2+X_3$, linearity of expectation and addition of variances for independent random variables give

The random variable $T$ is the sum $E+(-S)$. The linearity of expectation givesMultiplication by $-1$ reverses the mean but preserves the variance. Under the independence assumption used in part (b), the variance of a sum givesThus both predictions agree with the direct integrals. For the actual elapsed time, adding twenty-five changes the mean to twenty-five minutes and leaves the variance at $1/6$ minute squared.

Without independence, the correct expression is $\mathrm{Var}(T)=1/6-2\mathrm{Cov}(E,S)$ when both marginals are uniform. This explains why independent endpoint errors must be part of the intended model rather than a consequence of the display format.

The finite irreducible aperiodic Markov chain converges to its stationary distribution, regardless of its initial law. Therefore $\mathbb{E}[X_j]\to {\pi }_{+}-{\pi }_{-}$. Taking the arithmetic means of this convergent sequence givesThis uses linearity of expectation and convergence of Cesàro means; it does not require the variables $X_j$ to be independent.

Replacement and mixing make the one hundred draws independent and identically distributed. Writing $S={\sum }_{j=1}^{100}{X}_j$, linearity of expectation and additivity of variances of independent variables give

The finite irreducible aperiodic Markov chain converges to its stationary distribution, regardless of its initial law. Therefore $\mathbb{E}[X_j]\to {\pi }_{+}-{\pi }_{-}$. Taking the arithmetic means of this convergent sequence givesThis uses linearity of expectation and convergence of Cesàro means; it does not require the variables $X_j$ to be independent.

Generate independent and identically distributed random variables $X_1,\dots ,X_n$ with probability density function $p$, evaluate $\varphi (X_i)$ for each draw, and return the Monte Carlo estimatorLinearity of expectation makes this an unbiased estimator: $\mathbb{E}{\hat{\theta }}_n^{\mathrm{MC}}=\theta$. Independence and the specified variance give $\mathrm{Var}({\hat{\theta }}_n^{\mathrm{MC}})=1/n$. Its Monte Carlo standard error is therefore $n^{-1/2}$, and the strong law of large numbers gives convergence to $\theta$ almost surely.

Let $X_j$ be the indicator random variable that coin $j$ shows a head. Each expectation is $\mathbb{E}{X}_j=1/2$, so linearity of expectation gives the expected number of heads:For the game, write $p_i$ for the probability that $H$ eventually holds all $n$ coins when starting with $i$. The states $0,n$ are absorbing, so $p_0=0$ and $p_n=1$. Conditioning on the first toss and using the law of total probability givesThus $p_{i+1}-p_i=p_i-p_{i-1}$: successive differences are constant, and $p_i=A+Bi$. The boundary values give $A=0$, $B=1/n$, hence the gambler's ruin answer isFor completeness, the game terminates with probability one. From any nonabsorbing state, a run of $n$ consecutive heads would reach $n$ within the next $n$ tosses, and that run has probability $2^{-n}$. Conditional on survival, the same bound applies to the next block. Therefore the probability of surviving $kn$ tosses is at most $(1-2^{-n}{)}^k$, which tends to zero.

Let $I$ be the number of heads in the initial allocation tosses. Then $I$ has the binomial distribution with parameters $n$ and $1/2$. Given $I=i$, the winning probability is $i/n$, including the already finished games $i=0,n$. The law of total probability and linearity of expectation giveThus, for $n\ge 1$,

Assume the variances are finite. Expand the square and apply linearity of expectation:Part (i), applied to the identity functions, makes the bracketed covariance zero by independence of random variables. Hence

Part (a) with $n=6$ gives $\mathbb{E}[S_1]=\mathbb{E}[S_2]=7/2$ and $\mathrm{var}(S_1)=\mathrm{var}(S_2)=35/12$. By linearity of expectation,Independence of random variables and $\mathrm{var}(-S_2)=\mathrm{var}(S_2)$ give the same variance for the sum and difference:

By linearity of expectation and independence,For $\sigma >0$, apply Chebyshev's inequality at threshold $h\sigma$ and take the complementary event:If $\sigma =0$, every $X_i=\mu$ almost surely, so $W=\mu$ almost surely and the bound holds directly. This handles the zero-threshold case without applying Chebyshev's inequality at an inadmissible threshold.

For each random variable, the probability mass function givesLet $S_m={\sum }_{i=1}^m{X}_i$. By linearity of expectation and variance additivity for independent random variables,The independence is needed for the variance calculation, but not for the mean.

Both sides of (ii) are polynomials in the colour count, and they agree at every integer $x\ge 2$. A nonzero polynomial has only finitely many roots, so this is a polynomial identity and holds also at $x=0,1$. It can therefore be evaluated at every possible value of the Poisson random variable $X$.

Take expectations and use the factorial moments from (a)(iv). The sum over $k$ is finite, so linearity of expectation givesCombining this with (iii) proves Dobiński's formula for the Bell numbers:For completeness, the empty set has one partition of a set, giving $B(0)=1$; that case uses the empty-product conventions $0^0=1$ in this identity and a $k=0$ term in the partition sum.

Assume a finite second moment, so the following expectations are finite. Expand the square and use linearity of expectation:ThusThe same equality holds in the extended sense when the mean is finite and the second moment is infinite.

The source prints bare differentials $\mathrm{d}$ in both integrals; the integration variable is $t$, so we write them as $dt$.^[ref]

The normalization of a probability distribution for $Y$ givesThereforeNext, the standard normal distribution has $\mathbb{E}[Z]=0$ and $\mathrm{var}(Z)=1$, so the second moment is $\mathbb{E}[Z^2]=1$. Since $Y=Z^2/2$, linearity of expectation yields $\mathbb{E}[Y]=1/2$. Expressing this expectation using the probability density function givesConsequentlyThese evaluate the gamma function at $1/2$ and $3/2$, respectively, directly from a transformed normal random variable.

By linearity of expectation, $\mathbb{E}(S_n)={\sum }_{i=1}^{n}p=np$. Part (ii) gives $\mathrm{Var}(X_i)=\mathbb{E}(X_i^2)-[\mathbb{E}(X_i){]}^2=p-p^2$. Independent random variables have zero pairwise covariance, so the variance of their sum is the sum of their variances. Therefore