Past exam of the options II course of the University of Oxford / 2012 / M (Probability)

Each ticket is equally likely, so the supplied sums giveHenceRepeated numerical values still represent different equally likely tickets; they must retain their multiplicities.

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 event $S\ge 500$ is contained in $|S-400|\ge 100$. Thus Chebyshev's inequality givesThe variance bound is rigorous and uses no normal approximation.

The central limit theorem suggests approximating $S$ by a normal variable with mean $400$ and variance $560$. ThusSince $S$ is integer-valued, the usual continuity correction replaces $500$ by $499.5$:These are approximation formulas rather than consequences of Chebyshev's bound; the standard deviation is $\sqrt{560}$, not $560$.

Convergence in probability to $Y$ means that for every $\epsilon >0$,Here the variables must be defined on a common probability space so that their differences are meaningful.

Convergence in distribution to $Y$ means convergence of cumulative distribution functions at every continuity point of the limiting distribution function:Convergence at a jump of $F_Y$ is not required. Only the laws are involved, so this definition does not require a particular coupling of the variables.

Assume convergence in probability and fix $\epsilon >0$. On the event $|Y_n-Y|\le \epsilon$, $Y\le x-\epsilon$ implies $Y_n\le x$, while $Y_n\le x$ implies $Y\le x+\epsilon$. HenceTaking lower and upper limits in $n$ removes the error probabilities. If $x$ is a continuity point of $F_Y$, letting $\epsilon \downarrow 0$ squeezes both bounds to $F_Y(x)$. Thus

A homogeneous Poisson process of rate $\lambda >0$ is an integer-valued counting process with $N_0=0$ almost surely, nondecreasing right-continuous paths with unit jumps, and independent stationary increments satisfyingIncrements over disjoint time intervals are independent. In particular,The finite Poisson counts ensure that there is no explosion in finite time. The increment formula also gives stochastic continuity, since the probability of any arrival in an interval of length $h$ is $1-e^{-\lambda h}\to 0$.

Since $N_3\sim \mathrm{Poisson}(3\lambda )$,

Stationary increments give $N_5-N_3\sim \mathrm{Poisson}(2\lambda )$. ThereforeThere is zero probability of an arrival at either deterministic endpoint, so endpoint inclusion does not alter the result.

Successive interarrival times $W_1=T_1$ and $W_2=T_2-T_1$ are independent exponential variables of rate $\lambda$. Their joint density is ${\lambda }^2{e}^{-\lambda (w_1+w_2)}$ for $w_1,w_2>0$. The transformation $t_1=w_1$, $t_2=w_1+w_2$ has Jacobian determinant one, givingIntegrating over $0<t_1<t_2<\infty$ gives one, as a joint density should.

Write $N_3=N_2+(N_3-N_2)$. Independent increments make $N_2$ independent of the second term. HenceThe overlapping time interval contributes the covariance; the extra unit of time contributes none.

After the third arrival, the future interarrival times are independent of the first three interarrival times and again generate a rate-$\lambda$ Poisson process. Conditional on $T_3=t<2$, exactly three arrivals have occurred by time $t$, and the number of further arrivals before two is Poisson with mean $\lambda (2-t)$. Consequently, in the regular conditional-distribution sense at $t=1$,Explicitly its conditional mass is zero for $k<3$ and isThis conditions on the third arrival time itself, rather than merely on the event that the third arrival has happened by time one.