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Given $X=k$, the score $Y$ has a binomial distribution with success probability $1/2$. Its conditional expected value and variance are $k/2$ and $k/4$. ThusThis is binomial thinning of the first-round count; the second formula uses the identity $\mathbb{E}[Y^2\mid X]=\mathrm{Var}(Y\mid X)+\mathbb{E}[Y\mid X{]}^2$.

For $c>1$, the expected value is the convergent improper integralFor completeness, when $0<c\le 1$ the same positive integral diverges and the expected value is infinite.

Since $X\ge 0$, the function $X\mapsto X^2$ is strictly increasing on its support. For $y\ge 0$,and $F_Y(y)=0$ for $y<0$. Differentiating the cumulative distribution function givesThe density value at the single point $y=0$ is immaterial. Hence $Y$ has an exponential distribution with rate $1/2$. Direct integration by parts, or its standard expected value, yields

By the density formula for expected value,For the standard normal distribution, the integrand in the second moment is even, soComparing these two integrals gives

The law of the unconscious statistician gives the improper integralIn particular $\mathbb{E}(X)=1/2$ and $\mathbb{E}(X^2)=1/3$, soFor $p=-1$, ${\int }_{\epsilon }^1{x}^{-1}dx=-\log \epsilon \to +\infty$; for $p<-1$, the power integral also tends to $+\infty$. Thus $\mathbb{E}(X^p)=+\infty$ for $p\le -1$, so there is no finite expected value. Negative powers need only be defined almost surely on $X>0$; the null event $X=0$ has no effect on this conclusion.

First $\mu >0$. If a nonnegative random variable had zero expected value, it would equal zero almost surely, which is impossible for a distribution having a probability density function on $[0,\infty )$: a singleton has probability zero. Thus the given division by $\mu$ is legitimate.

Because $x\ge 0$ and $f(x)\ge 0$, $g(x)=xf(x)/\mu \ge 0$. MoreoverExtend $g$ by zero for negative $x$. It is a probability density function, defining the size-biased distribution of $X$: larger observations receive weight proportional to their value.

The expected value is the derivative of the probability generating function at one. The expression from part (ii) givesIts finiteness also follows directly from the exponentially decaying geometric probabilities.

Since $Y\ge 1$, its finite expected value satisfies $1\le \mu <\infty$. Every $q_k=k{p}_k/\mu$ is nonnegative andThus these values are a probability mass function. They define the size-biased distribution of $Y$: larger values receive proportionally more weight.

Because $T$ and $U$ are independent random variables, their product's expected value factors:The original independence assumption is essential to this factorization.

Normalization gives $c=8/3$. Integrating the probability density from one to two givesThe expected value exists, since the first-moment integrals converge, and equals

For $q>1$, decreasing the second parameter by one gives another valid beta distribution density andNormalization therefore givesThe restriction $q>1$ is essential for this finite expected value; when $q=1$, the singularity near $1$ instead makes the expectation infinite.

By independence, the joint probability density is the product of the two uniform distribution densities:Changing its values on the boundary does not affect any probability or expected value.

Differentiate the probability generating function equation:At $s=1$, $G_N(1)=1$, givingThe derivative is finite because $p<1/2$ and the explicit square root has no singularity at $s=1$. Alternatively, the expected population in generation $t$ is $(2p{)}^t$, so the expected number of splits in that generation is $p(2p{)}^t$. Summing this nonnegative geometric series gives the same answer without initially assuming a finite expected value.

Markov's inequality: if $X\ge 0$ and $a>0$, then $\mathbb{P}(X\ge a)\le \mathbb{E}[X]/a$. Pointwise,Taking expected values, using their monotonicity and the identity $\mathbb{E}[1_E]=\mathbb{P}(E)$, gives $\mathbb{E}[X]\ge a\mathbb{P}(X\ge a)$. Division by the positive threshold proves the inequality. If the expectation is infinite, the extended-valued bound remains true but uninformative.

A general-purpose high-dimensional choice is Monte Carlo integration. Draw independent vectors $W_r$ uniformly from the unit cube and evaluate $H(W_r)={\sum }_{i=1}^{100}{e}^{\Vert W_r-a_i{\Vert }^2}$. The cube has volume $1$, so its integral is the expected value of $H(W_1)$. The iterative estimator isThe integrand is continuous on a compact set, hence bounded with finite variance. The strong law of large numbers gives almost sure convergence, and the square root of the mean squared error is exactly $\sigma /\sqrt{N}$, where ${\sigma }^2=\mathrm{Var}(H(W_1))$. Thus the standard stochastic accuracy rate is $O(N^{-1/2})$; the dimension affects the variance and evaluation cost, not this exponent.

There is also an important structure-specific improvement here: $e^{\Vert x-a_i{\Vert }^2}={\prod }_{j=1}^{100}{e}^{(x_j-a_{ij}{)}^2}$ is a separable integrand. The Fubini theorem reduces the requested value exactly toOne can therefore replace high-dimensional sampling by $10000$ one-dimensional composite Simpson's rule calculations. For each factor use even $n$, $h=1/n$, andthen iterate $n\leftarrow 2n$ and form the sum of the products. These smooth factors give $O(n^{-4})$ deterministic discretization error for fixed centres. Thus Monte Carlo integration is a valid generic answer, while exploiting separability is the more efficient specialized alternative; a large dimension alone does not force a tensor-product grid.

Let $A_i$ be the event that candidate $i$ succeeds. The countable additivity axiom adds probabilities only for mutually disjoint events. Successes of different candidates are not mutually exclusive, so replacing the probability of their union by ${\sum }_i\mathbb{P}(A_i)=1$ misuses that axiom. Without disjointness one obtains only the union bound, not certainty. The same sum is the expected value of the number of successes, which also need not imply at least one.

If the ten outcomes are mutually independent events, with each success probability $0.1$, all fail with probability $0.9^{10}$. HenceThe independence assumption is essential. The marginal probabilities alone allow any union probability from $0.1$ up to $1$; perfectly overlapping success events give the lower endpoint and disjoint successes give the upper endpoint.

Because $s>1$, the function $t\mapsto s^t$ is increasing and nonnegative. Thus $\{X\ge x\}=\{s^X\ge s^x\}$. Applying Markov's inequality to the nonnegative random variable $s^X$, whose expected value is finite by assumption, yieldsUnder the usual probability generating function convention $X$ takes nonnegative integer values. The argument itself works for any real random variable if $G_X(s)$ is understood as the finite expected value $\mathbb{E}[s^X]$.

Choose $g(t)=e^t$. If $W$ has an exponential distribution with rate $r$, then for $c<r$,while the integral diverges to $+\infty$ when $c\ge r$. Since $Z$ has rate $3$,For the other two rates, $\mathbb{E}[g(V)]=2/(2-1)=2$, whereas $\mathbb{E}[g(U)]={\int }_0^{\infty }1dt=+\infty$. These computations establish exactly the required finite variance, finite expected value, and infinite expected value.

Write $h_i={\mathbb{E}}_i[J]=i/k$ and $v_i={\mathbb{E}}_i[JX]$. Since $0\le JX\le X$, the latter expected value is finite. Conditional on the first move to $i-1$ or $i+1$, total time is one plus the remaining time, while the terminal success indicator random variable is unchanged. The Markov property and linearity of expectation giveAt either absorbing boundary the remaining time is zero, so $v_0=v_k=0$.

Let $w_i=i(k^2-i^2)/(3k)$. The linear term has zero second difference and $i^3$ has second difference $6i$. Consequentlywhich is exactly the required recurrence, and $w_0=w_k=0$. Uniqueness follows as before: the difference of two solutions has zero second difference and zero endpoint values. HenceThis is the success-weighted duration underlying the duration of gambler's ruin conditioned on a chosen boundary.

Let $t_i={\mathbb{E}}_i[X]$. Time is accumulated in each transient chamber before the next move; arrival at either terminal state ends the trial immediately. Thus $t_0=t_k=0$. A block of $k$ identical rightward choices forces absorption, so the geometric tail bound from a uniform escape probability gives finite expected value of $X$.

First-step analysis now givesThe quadratic $t_i=i(k-i)$ has the correct boundary values and second difference $t_{i+1}-2t_i+t_{i-1}=-2$, so it solves this system. The difference of two solutions would be linear in $i$ with zero values at both endpoints, hence zero. ThereforeThis is the expected duration of symmetric gambler's ruin. It includes the first minute in the starting transient chamber and does not add a minute after absorption.