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For an integrable real random variable $X$, its cumulant-generating function is the extended-real convex functionwhere $\psi (\theta )=+\infty$ if the exponential moment diverges. Its Legendre transform of a cumulant-generating function is

For a centered random variable $X$, being sub-Poisson in the right tail with variance parameter ${\sigma }^2$ meansIt is sub-Gamma in the right tail with variance parameter $v$ and scale parameter $c$ when

A centered random variable $Y$ is sub-Gaussian with variance parameter ${\sigma }^2$ whenThe Chernoff bound, applied to $Y$ and $-Y$, yieldsThe tail integral formula for moments and the substitution $u=t^2/(2{\sigma }^2)$ now give

For an integrable random variable $X$, the law of total expectation for the partition of a sample space isParts of zero probability contribute zero and need no definition of $\mathbb{E}[X\mid B_i]$. To see the identity, write $X={\sum }_{i}X{1}_{B_i}$ and use linearity of expectation; integrability justifies the countable sum since ${\sum }_i\mathbb{E}[|X|1_{B_i}]=\mathbb{E}[|X|]<\infty$.

Each random variable has mean zero and second moment $2/3$, so its variance is $2/3$. For $S_m={\sum }_{i=1}^m{X}_i$, linearity of expectation and variance additivity for independent random variables give

For a nonnegative integer-valued random variable, its probability generating function isThe convention $s^0=1$, including at $s=0$, gives $G(0)=p(0)$. The power series has absolute convergence for every real $|s|\le 1$, since ${\sum }_{j}p(j)|s{|}^j\le 1$. More generally its real domain as a finite expected value is $\{s\in \mathbb{R}:{\sum }_{j}p(j)|s{|}^j<\infty \}$, which may extend beyond $[-1,1]$. No extension beyond that guaranteed interval is automatic.

Every assignment determines a unique set partition into its nonempty colour classes. If there are $k$ such blocks of a set partition, there are $A(n,k)$ choices for that set partition. For each fixed choice, assign distinct colours to the $k$ blocks of a set partition inways. Although the blocks of a set partition are unlabelled, they are distinguishable by their elements: ordering them temporarily, for instance by their least elements, counts each assignment exactly once. The falling factorial is zero when $k>x$, as it should be. Summing over the possible numbers of used colours provesHere $A(n,k)$ is the Stirling number of the second kind. Both sides are polynomials in $x$ and agree at infinitely many integers $x>2$, so the identity also holds for every real $x$, including $0,1,2$. This extension will justify substituting the random variable $X$ in part (iii).

Expand the square and use linearity of expectation, together with $\mathbb{E}[X]=\mu$ and $\mathbb{E}[1]=1$:This proof applies whenever the second moment is finite, and does not require the random variable to be continuous.

For a random variable with a continuous probability distribution and probability density function $f_X$, the mean and variance areA finite mean requires $\int |x|f_X(x)dx<\infty$. A finite variance additionally requires the second moment to be finite; without that condition the nonnegative integral for the variance may be infinite.

For random variables with finite second moments, the variance of a sum isNo independence assumption is needed.

For a random variable with a probability density function, a singleton has zero probability:Equivalently, its cumulative distribution function is continuous, so it has no jump at $x$.

For a nonnegative integer-valued random variable $W$, the tail-sum formula states $\mathbb{E}[W]={\sum }_{k=1}^{\infty }\mathbb{P}(W\ge k)$. It follows by writing $W={\sum }_{k\ge 1}{1}_{\{W\ge k\}}$ and using nonnegativity to interchange sum and expectation.

With $W=\lfloor X\rfloor$, this givesThe Riemann zeta function is finite here because $t-1>1$.

For an outcome $\omega =(i,j)$, the two random variables are the coordinate mapsTheir marginal probability mass functions are $p_X(i)=p_Y(j)=1/6$ on $\{1,\dots ,6\}$, and their joint probability mass function satisfiesOff that support, the same factorization holds with zeros. Hence $X$ and $Y$ are independent random variables, by the joint probability mass function criterion for independence.

For random variables with finite second moments, their covariance isThe Cauchy-Schwarz inequality ensures the product is integrable. Expand and use linearity of expectation:The same identity holds whenever the quantities involved exist with sufficient integrability.

An expected value of a signed random variable is finite precisely when its absolute value is integrable. With the standard normal distribution probability density function,For $\alpha <1/2$, this integral equals $1/[\sqrt{2\pi }(1/2-\alpha )]<\infty$. For $\alpha \ge 1/2$, its integrand does not decay sufficiently and the integral diverges. ThereforeThe value is zero by the odd function and the symmetric distribution. For the other $\alpha$, both positive and negative parts have infinite integrals; formal cancellation would give only a Cauchy principal value, not an expected value. This distinction is the integrability threshold for exponential powers of a Gaussian.

For a continuous random variable with probability density function $p$, its expected value and variance areThe mean requires an integrable first absolute moment, and a finite variance requires a finite second moment. When these hold, $\mathrm{Var}X=\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$.

Interpreting $f$ as the probability density function of the random variable, for $\alpha \le \beta$ one hasA single point has zero probability, so including or excluding either endpoint does not change this result.

For a continuous random variable, the probability density integrates to the probability of an interval:Individual endpoints have zero probability, so strict or non-strict inequalities give the same answer.

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 preceding marginal calculation makes each $V_i$ uniformly distributed on $(0,1)$, with expectation $1/2$ and variance $1/12$. By independence,Apply part (a) to the centred random variable $Z=S_n/n-1/2$ to obtainThis is Chebyshev's inequality in this setting.