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Let $N=T-t$ and use the discounted multipliers $u,d$. Under $Q$, the number $K$ of up moves in the remaining $N$ periods has the binomial distribution $\mathrm{Binomial}(N,q)$. Sincethe conditional expectation in part (b) becomesFor an undiscounted stock, replace $u,d$ by $U,D$ and discount the resulting payoff by the appropriate power of $R$.

The Chebyshev inequality states that for a random variable $Y$ with finite variance and every $\delta >0$,Indeed, Markov inequality applied to the nonnegative variable $(Y-\mathbb{E}Y{)}^2$ gives

The sum $S_n=X_1+\cdots +X_n$ has the binomial distribution, soFor fixed $n$ this is a finite sum of polynomial functions of $p$, hence is a polynomial.

Here $X$ has the binomial distribution $\mathrm{Bin}(n,\theta )$, so, up to a factor independent of $\theta$, the likelihood function isFor $0<X<n$, differentiating the log-likelihood givesand the boundary cases give the same formula. Thus the binomial proportion maximum-likelihood estimator is

The number of heads has the binomial distribution. Choosing the $k$ head positions gives

Under a risk-neutral measure, the discounted stock must be a martingale. If $q$ is the probability of the factor $1+b$, this condition issoThe hypothesis $a<r<b$ ensures $0<q<1$, making this equivalent to the original probability measure. Since the $N$ returns are independent under $Q$, the number of up moves has a binomial distribution, and hence

For fixed disjoint $A,B$, the random variable $e(A,B)$ has the binomial distribution $\mathrm{Bin}(|A||B|,p)$. A two-sided Chernoff bound givesWhen $|A|,|B|\ge n/\log n$, this is at mostThere are at most $3^n$ ordered disjoint pairs $(A,B)$, since each vertex can lie in $A$, in $B$, or in neither. The union bound therefore makes the probability of any failure at mostwhich proves the simultaneous estimate.

LetHere $S_n$ has the binomial distribution with parameters $(n,\lambda /n)$. Part a, now with independent coordinates, givesBy Pinsker's inequality, the probability mass functions therefore converge in total variation, and in particular $S_n$ converges in distribution to $Z\sim \mathrm{Poisson}(\lambda )$.

The joint probability of the observed row depends only on $S_n$:Taking logarithms in any fixed base and choosing $c_n=\log n$ giveswhereThe convergence lemma supplied in the question now yieldsThis sparse triangular array therefore has a random limiting normalized self-information rather than the constant limit in the usual asymptotic equipartition property.

Explore the connected component of a graph containing a fixed vertex $v$ by breadth-first search. If the exploration discovers at least $m$ vertices, then before its $m$th discovery at least $m-1$ of at most $mn$ tested potential edges must be present. The tests are independent Bernoulli trials with parameter $p=(1-\epsilon )/n$, soThe binomial distribution on the right has mean $(1-\epsilon )m$. For $m\ge 2/\epsilon$, the exponential Markov bound givesfor an absolute constant $c>0$. The union bound over the $n$ choices of $v$ now givesTaking $m=C_{\epsilon }\log n$ with $C_{\epsilon }>2/(c{\epsilon }^2)$ makes this probability tend to zero. This proves the subcritical component bound for a binomial random graph.

Under the boundary null $p={\pi }_0=0.20$, rejection requires both $X_1>r_1$ and $X_1+X_2\ge r$. Independence of the two binomial distributions givesThe rejection event is increasing in $p$, so this boundary value is the supremum over the composite null $p\le {\pi }_0$.

Take independent Bernoulli random variables $X_i$ of success probability $r$, independently of $N$. On the event $N=n$, the random sum ${\sum }_{i=1}^N{X}_i$ has the binomial distribution with parameters $(n,r)$, including the empty sum zero at $n=0$. It therefore has exactly the specified joint distribution with $N$.

Thus one may realize $\boxed{Z=\sum _{i=1}^N{X}_i}$ on a suitable probability space. This is a distributional construction; it does not require a pre-existing $Z$ to be independent of $N$.

The number $R$ of successes has a binomial distribution. Each arrangement of $r$ successes and $n-r$ failures has probability $p^r{q}^{n-r}$ by independence, and the binomial coefficient counts the arrangements. ThereforeThe final equality is the binomial theorem. Writing $R$ as a sum of independent Bernoulli random variables gives its expectation and variance: $\mathbb{E}R=np$, $\mathrm{Var}(R)=npq$. Thus the matching normal distribution hasThe normal approximation is appropriate when both expected counts $np$ and $nq$ are sufficiently large; large $n$ alone is insufficient if $p$ is extremely close to zero or one.

For the guessing student, $n=60$, $p=1/4$, $q=3/4$. Therefore the expected number of correct answers is $\boxed{15}$, and ${\sigma }^2=45/4$, $\sigma =3\sqrt{5}/2$. The inclusive integer range $13,14,15,16,17$ becomes $12.5\le X\le 17.5$ after the continuity correction. Standardizing the normal distribution givesHere $(5/2)/(3\sqrt{5}/2)=\sqrt{5}/3$, and the last step uses the supplied bound. The approximate probability is $0.54394$. The normal approximation provides the requested conclusion; as a separate check, summing the five exact binomial probabilities gives approximately $0.54370$, also exceeding one half.

Use the intended model of independent Bernoulli trials, with the same scoring probability $p$ for every shot. Equal marginal scoring probabilities alone would not establish the stated formula without independence. Let $N_1,N_2$ count misses in the two groups of five. They are independent random variables with binomial distribution $\mathrm{Bin}(5,1-p)$. A tie occurs precisely when $N_1=N_2=n$ for one of the mutually exclusive possibilities $n=0,\dots ,5$. Therefore the tie probability for independent binomial counts isAt $p=1/2$, Vandermonde's identity gives ${\sum }_n{\left(\genfrac{}{}{0ex}{}{5}{n}\right)}^2=\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$, soExchanging the two teams preserves the distribution and interchanges the two winning events. Their total probability is $1-\alpha$, so the first team wins at the end of the initial period with probability $193/512$. In this model the shooting order gives no advantage.

Assuming the egg colours are independent with the same probability $p$, the number of white eggs has the binomial distribution with four trials. The possibilities are

The binomial coefficients count which eggs are white. Marginal colour probabilities alone would not determine nest probabilities without the independence assumption.

The exponential series proves normalization of a probability distribution:Let $M$ and $F$ count the two labels. Conditional on $N=M+F=k+l$, independent labeling gives a binomial distribution, so the joint probability mass function isEach bracket is a normalized Poisson distribution. Summing over $l\ge 0$ therefore givesThe factorization also proves that $M$ and $F$ are independent random variables. This is Poisson thinning; the phrase about ignoring the other count refers to taking a marginal distribution, rather than fixing that count.

For the binomial distribution and $T\ge 1$, the identity $s\left(\genfrac{}{}{0ex}{}{T}{s}\right)=T\left(\genfrac{}{}{0ex}{}{T-1}{s-1}\right)$ gives the expectationThe final sum is one by the binomial theorem. For $T=0$ the variable is identically zero, so the same answer holds:

The lecture attendances are independent Bernoulli trials with success probability $f$. Their count has the binomial distribution:

Neither biscuit type can run out in this case. Treat each freely made choice as an independent Bernoulli trial, with success meaning chocolate. The number taken after $n$ choices has a binomial distribution, soThe binomial coefficient counts the possible success positions, and every such arrangement has the same probability. Independence of successive choices is the probabilistic model implicit in the question; identical marginal preferences alone would not imply it.

A forced plain choice occurs exactly when the last chocolate is taken before the final Fellow is served: if it is taken by choice $f\le F-1$, choice $f+1$ has no chocolate available, whatever that person's preference would have been. Exhaustion precisely on choice $F$ forces nobody. The distinct possible exhaustion times are disjoint events, henceEquivalently, imagine independent preferred choices even after stock is exhausted. The event is that at least $C$ preferred chocolate choices occur among the first $F-1$ people, so the binomial distribution gives $P_{\mathrm{Disaster}}={\sum }_{k=C}^{F-1}\left(\genfrac{}{}{0ex}{}{F-1}{k}\right)p^k(1-p{)}^{F-1-k}$. The success-count and waiting-time duality explains equality of these two sums. Plain-biscuit depletion cannot interfere because $D>F$.

After $j$ choices, let $C_j$ be the number of forward moves. The two unwrapped positions are $C_j$ and $F-(j-C_j)$, whose gap is $F-j$. It follows that the first subsequent meeting occurs exactly at step $F$, regardless of the order of the choices. Its position is the number of forward moves in that block.

That count has a binomial distribution with parameters $F,p$. ThereforeThe two endpoint labels $0$ and $F$ represent the same physical seat reached by different complete circuits; the unwrapped convention distinguishes their probabilities.

Write the count as a sum of $24$ Bernoulli random variables, one for each attendance indicator. Each indicator random variable has expected value $3/4$, so linearity of expectation givesEquivalently, $N$ has a binomial distribution with parameters $24$ and $3/4$.