Since $Y=Z^2/2$ is nonnegative, its cumulative distribution function is zero for $y<0$. For $y\ge 0$,The last equality uses the symmetry of the standard normal distribution. Thus

The second moment isUsing the mean $\mathbb{E}[X]=1$, the variance is therefore

Normalization of the probability density function and the specified mean give two equations:Dividing the second by the first gives $2b/3=1$. ConsequentlyBoth are positive, as required for the stated probability density function.

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.

Computing the same expected value directly from the Poisson distribution instead givesFor each fixed positive integer $n$, the ratio test shows that this sum converges. This identity is Dobiński's formula. The case $n=0$ also gives $B(0)=1$ if $0^0$ is interpreted as one in this particular moment formula.

Substitute $X$ into the polynomial identity and apply linearity of expectation to its finite sum. The factorial moments of the Poisson distribution in part (a)(iv) are all one, givingThus its $n$th moment counts set partitions: it equals the $n$th Bell number.

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).

Each of the $n$ distinct elements can receive any of the $x$ labelled colours. The multiplication principle therefore givescolour assignments; colours are allowed to remain unused.

For the specified Poisson distribution, the exponential series gives the probability generating functionEvery derivative of $G$ is again $e^{s-1}$, so the factorial moments and variance areAlso $\mathbb{E}[X]=1$.

Since $X^2=X(X-1)+X$, the second moment is $\mathbb{E}[X^2]=G^{\prime \prime }(1)+G^{\prime }(1)$. Subtracting the squared mean gives

Under the extra hypothesis that $G(s)$ exists for every real $s$, its power series has infinite radius of convergence. Termwise differentiation of a power series is therefore valid to every finite order. For $k\ge 1$,At $s=1$ this is the expected value of the falling factorial of $X$:In particular $\boxed{G^{\prime }(1)=\mathbb{E}[X]}$. The terms with $j<k$ vanish because their falling factorials contain a zero factor. These factorial moments are finite under the stated hypothesis.

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.

Each $X_i$ has the same probability distribution as $-X_i$, and independence makes $S_m$ and $-S_m$ identically distributed. Thus their two strict tail probabilities are equal:Apply Chebyshev's inequality using the mean and variance from part (i):The equal-tail argument is essential: applying Chebyshev's inequality directly to the absolute tail alone gives the weaker bound $8/(3m)$.

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

At an absorbing state, the remaining absorption time is zero, so $e_0=e_n=0$. Multiplying the linear recurrence relation by three givesThe second finite difference of $k^2$ is two. A particular solution is therefore $-3k^2/2$, while every solution of the corresponding homogeneous linear recurrence relation has the form $A+Bk$. The boundary conditions give $A=0$ and $B=3n/2$, henceThis is the mean absorption time of a lazy symmetric random walk with holding probability $1/3$.

First the expected value of the absorption time is finite. From any interior state, consecutive leftward moves lead to absorption within at most $n$ steps, with probability at least $p=3^{-n}$. By independence of the subsequent steps, the probability of surviving $jn$ steps is at most $(1-p{)}^j$. Summing this geometric series bounds the expected value of the absorption time by $n/p$.

Now apply first-step analysis. Starting from $k$ in the interior, one step is always used. Conditional on that step, the remaining expected values are $e_{k+1}$, $e_k$, and $e_{k-1}$, each with probability $1/3$. The law of total expectation therefore givesThe term $1$ remains even when the particle stays at $k$, because a step has elapsed. Rearranging gives

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 ${\sum }_i\mathbb{E}[|X|1_{B_i}]=\mathbb{E}[|X|]<\infty$.

The events $B_1,B_2,\dots$ form a partition of a sample space $\Omega$ when they are pairwise disjoint sets and cover the whole sample space:Thus exactly one of these events occurs on every outcome. Under the usual set partition convention the parts are nonempty; empty parts can simply be discarded.

Choose $h_n=\min \{1/n,(b-c)/2\}>0$. Then $h_n\to 0$ and $[c,c+h_n]\subset (a,b)$. The mean value theorem states that a continuous function on a closed interval, differentiable in its interior, has an interior derivative equal to its endpoint difference quotient. It gives $x_n\in (c,c+h_n)$ withHere $x_n\ne c$, $x_n\to c$, and the right-hand side tends to $f^{\prime }(c)$ by differentiability at $c$. If $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, the same values $f^{\prime }(x_n)$ must tend to $L$. Uniqueness of the limit of a sequence givesThus a derivative cannot have a removable discontinuity at a point where the original function is differentiable. The proof also excludes an infinite limit of a function there, since the displayed difference quotients have a finite limit of a sequence.