Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 2 27I a Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 2 27I c Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 4 29I Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2011 ia Paper 2 3F b Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2011 ia Paper 2 3F c Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 1 29J i Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 2 11F iii Solution Created 2026-09-24 Updated 2026-10-07
For , substitute in the probability generating function. The moment-generating function of isThe domain follows from convergence of the generating series; at or above that positive threshold the expectation diverges, even though a rational continuation can be written.
Conditioning on survival, the probability mass function from part (ii) becomes a positive-integer geometric distribution:Its conditional moment-generating function is , tending to for fixed . We can prove the required limit directly, without invoking a continuity theorem. Set . For every , the strict upper tail is exactlySince and ,At both tails equal one. This is the exponential limit of surviving geometric branching. Conditioning is essential: the unconditional variable has probability of being zero, whereas the rare surviving population has size of order .
Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 2 9F iii Solution Created 2026-09-24 Updated 2026-10-07
Let and . Since they are independent, applying the product rule for moment-generating functions to givesAll factors are finite and strictly positive for every real by the stated hypothesis. Applying the same identity to and dividing yields , orThe centered unit-variance Taylor expansion from part (ii) gives and the same expansion for . Consequently .
For the first step of dyadic rigidity of a moment-generating function, take logarithms, which are allowed because . For each fixed and every integer ,Since , the right side tends to zero: it is . Thus for every , and the moment-generating function is even.
The original identity now becomes . Iterating this time with the correct fourth-power scaling givesNear zero, , so the right side tends to . ThereforeThis is the moment-generating function of a standard normal variable. By the uniqueness theorem for moment-generating functions, each of and has the standard normal distribution. The proof establishes the Gaussian characterization by independent sum and difference directly from the two functional identities and the first two moments.
Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 2 9F ii Solution Created 2026-09-24 Updated 2026-10-07
Finiteness of the moment-generating function on an open neighborhood of zero gives its differentiability there. Using the permitted interchange of differentiation and expectation,Hence , and . The second-order Taylor expansion with its Peano remainder givesHere is the raw second moment, not the variance: . This distinction matters before specializing to a centered variable.
Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 2 9F i Solution Created 2026-09-24 Updated 2026-10-07
The moment-generating function is at real for which that expectation is finite; . For independent random variables, the joint law factorizes, soThis is the transform form of independence. Its ordinary finite-valued interpretation applies where both factors are finite, with the transform of a zero multiple equal to .
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 32 4 Solution Created 2026-10-03 Updated 2026-10-07
An effective bandwidth converts a traffic distribution and a quality-of-service requirement into a capacity requirement, penalizing bursts as well as mean load. Let be the traffic generated in one observation interval and assume its moment-generating function is finite for some positive parameters. Write . The Chernoff bound follows immediately from Markov's inequality applied to :Optimizing gives , where the supremum is restricted to finite exponential moments. This exhibits the effective bandwidth as the load appropriate to exponential tail control, not an extra physical stream of traffic.
For the mean and peak limits of effective bandwidth, if exponential moments exist near zero, the cumulant-generating function expansion isFor small , , with the variance giving the first burstiness correction. convexity of and show that its secant slope is nondecreasing; Jensen's inequality also gives .
If is bounded with essential supremum , then . For any , the probability is positive andLet and then . For large , , so very stringent exponential tail control approaches peak provisioning. If traffic is unbounded and every positive exponential moment is finite, the same lower-bound argument gives . If exponential moments cease to exist beyond a finite parameter, is infinite there; one cannot assign a finite large- peak interpretation to that traffic.
For example, deterministic traffic has . A burst of size with Bernoulli distribution probability giveswhich moves from to . For Poisson packet count of mean and packet size , , showing a mean limit but no finite peak limit.
For independent traffic sources, logarithmic generating functions add, so effective bandwidths add. For arrivals in a window of length , use the rate-valued definition . A resource with rate capacity then satisfiesThus a sufficient admission condition for a target probability is for some . Independence is needed for this additive form; correlated sources require the joint exponential moment.
A queue involves a supremum over time windows, not merely one fixed window. For independent identically distributed discrete-time traffic increments , constant service , and stationary workloadthe workload Chernoff bound for independent increments follows from a union bound:, whenever ,The requirement is precisely . This explains why effective bandwidth is useful for buffer and delay design: the relevant time scale and tail parameter depend on the service rule and target overflow risk. A single-window bound cannot be silently substituted for an infinite-horizon workload bound.