Quantitative finance often boils down to parameter estimation: given some model which you believe to be true, what are the parameters for that model that fit the actual data?
One type of parameter estimation is maximum likelihood estimation. The MLE of a distribution given some data has an intuitive definition: it is the parameter(s) for the model that would make observing the given data most likely.
What is the MLE for the success probability of a geometric distribution which takes an observed \(n=4\) trials to achieve the first success?
You have a coin which you know lands on heads with probability \(P.\) If it is flipped 10 times and there are 6 heads, what is the MLE for \(P?\)
In the last question, you might think “that seems like a bad estimate, since most coins are essentially fair”. This is where having a prior distribution of beliefs (“a priori”) can be extremely helpful.
You have a coin which you know lands on heads with probability \(P.\) You believe that P is normally distributed with mean 0.5 and variance 0.01. If it is flipped 10 times and there are 6 heads, which is the best estimate for \(P?\)
Technically speaking, we're looking for a "maximum a posteriori probability" (MAP). It's like an MLE, except that we Bayesian update from a non-uniform prior distribution for the parameter.
Also, \(P\) is a probability, which means that it must be bounded between 0 and 1. Assuming a normal distribution for \(P\) assigns a nonzero probability of finding a value for \(P\) outside of the interval \( [0,1].\) This assumption is only an approximation, which is very good for appropriate choices of mean and variance.
An airline has numbered their planes \(1,2,\ldots,N,\) and you observe the following 3 planes, which are randomly sampled from the \(N\) planes:
What is the maximum likelihood estimate for \(N?\) In other words, what value of \(N\) would, according to conditional probability, make your observation most likely?
In the previous problem, we saw an example where our estimator seemed poor and we would have preferred the unbiased estimator: one which has an expected value equal to the true value of the parameter.
When calculating a “sample variance,” you divide by a different constant instead of \(n\) in order to get an unbiased estimator. What is that constant?