Over the past week, I’ve given several tutorials to second year undergraduates preparing for upcoming papers on probability and statistics. In particular, I’ve now seen a lot of solutions to a lot of past papers and specimen questions, and it’s worthwhile to consider some of the typical mistakes students can make on these questions. Of course, as with any maths exam, there’s always the possibility of a particularly subtle or involved question coming up, but if the following three common areas of difficulty can be avoided, you’re on track for doing well.
In a previous course, a student will learn how to calculate the pdf of a function of a random variable. Here, we move onto the more interesting and useful case of finding the (joint) density of function(s) of two or more random variables. The key thing to remember here is that manipulating pdfs is not a strange arbitrary exercise – it is just integration. It is rarely of interest to consider the value of a pdf at a single point. We can draw meaningful conclusions from a pdf or from comparison of two pdfs by integrating them.
Then the question of substituting for new random variables is precisely integration by substitution, which we are totally happy with in the one-dimensional case, and should be fairly happy with in the two-dimensional case. To get from one joint density to another, we multiply by the absolute value of the Jacobian. To ensure you get it right, it makes sense to write out the informal infinitesimal relation
This is certainly relevant if we put integral signs in front of both sides, and explains why you obtain rather than the other way round. Note though that if is easier to calculate for some reason, then you can evaluate this and take the inverse, as your functions will almost certainly be locally bijective almost everywhere.
It is important to take the modulus of the Jacobian, since densities cannot be negative! If this looks like a fudge, then consider the situation in one dimension. If we substitute for , then f’ is obviously negative, BUT we also end up reversing the order of the bounds of the integral, eg [1/3, ¾] will become [2/3,1/4]. So we have a negative integrand (after multiplying by f'(x)) but bounds in the ‘wrong’ order. These two factors of -1 will obviously cancel, so it suffices just to multiply by |f'(x)| at that stage. It is harder to express in words, but a similar relation works for the Jacobian substitution.
You also need to check where the new joint density is non-zero. Suppose X, Y are supported on [0,1], then when we write we should indicate that it is 0 off this region, either by splitting into cases, or adding the indicator function as a factor. This is even more important after substitutions, as the range of the resulting random variables might be less obvious than the originals. Eg with X,Y as above, and , the resulting pdf will be non-zero only when . Failing to account for this will often lead to ludicrous answers. A general rule is that you can always check that any distribution you’ve produced does actually integrate to one.
Convergence using MGFs
There are two main reasons to use MGFs and PGFs. The first is that they behave nicely when applied to (possibly random) sums of independent random variables. The independence property is crucial to allow splitting of the MGF of the sum into the product of MGFs of the summands. Of course, implicit in this argument is that MGFs determine distributions.
A key theorem of the course is that this works even in the limit, so you can use MGFs to show convergence in distribution of a family of distributions. For this, you need to show that the MGFs converge pointwise on some interval [-a,a] around 0. (Note that the moments of the distribution are given by the family of derivatives at 0, as motivation for why this condition might be necessary.) Normally for such questions, you will have been asked to define the MGF earlier in the question, and probably will have found the MGF of a particular distribution or family of distributions, which might well end up appearing as the final answer.
Sometimes such an argument might involve substituting in something unusual, like t/N, rather than t, into a known MGF. Normally a Taylor series can be used to show the final convergence result. If you have a fraction, try to cancel terms so that you only have to evaluate one Taylor series, rather than lots.
Using the Markov Property
The Markov property is initially confusing, but once we become comfortable with the statement, it is increasingly irritating to have to answer the question: “show that this process has the Markov property.” This question is irritating because in most cases we want to answer: “because it obviously does!” Which is compelling, but unlikely to be considered satisfactory in a mathematics exam. Normally we observe that the random dynamics of the next step are a function only of the present location. Looking for the word ‘independent’ in the statement of the process under discussion is a good place to start for any argument along these lines.
The most developed example of a Markov process in this course is the Poisson process. I’ve written far too much about this before, so I won’t do so again, except to say this. When we think of the Poisson process, we generally have two thoughts going through our minds, namely the equivalent definitions of IID exponential inter-arrival times, and stationary, Poisson increments (or the infinitesimal version). If we draw a sketch of a sample trajectory of this process, we can label everything up and it is clear how it all fits together. But if you are asked to give a definition of the Poisson process , it is inappropriate to talk about inter-arrival times unless you define them in terms of , since that is the process you are actually trying to define! It is fine to write out
but the relation between the two characterisations of the process is not obvious. That is why it is a theorem of the course.
We have to be particularly careful of the difference in definition when we are calculating probabilities of various events. A classic example is this. Find the distribution of , conditional on . It’s very tempting to come up with some massive waffle to argue that the answer is 3+Po(1). The most streamlined observation is that the problem is easy if we are conditioning instead on . We just use the independent Poisson increments definition of , with no reference to inter-arrival times required. But then the Markov property applied at time 1 says that the distribution of depends only on the value of , not on the process on the interval [0,1). In a sense, the condition that is giving us extra information on the behaviour of the process up to time 1, and the Markov property, which we know holds for the Poisson process, asserts precisely that the extra information doesn’t matter.