What is the Probability Distribution Function of \(Z_1\)?

[CaptainBlack]

I was browsing the 1998 STEP paper 3 (like you do) and come across a question related to a real intelligence gathering problem.

It has some simplifications and probably invalid assumptions, but is still quite interesting.

The question is:

A hostile naval power possesses a large, unknown number \(N\) of submarines. Interception of radio signals yields a small number \(n\) of their identification numbers \(X_i, \ \ i = 1, 2,..., n\), which are taken to be independent and uniformly distributed over the continuous range from \(0\) to \(N\). Show that \(Z_1\) and \(Z_2\), defined by\( \displaystyle Z_1 = \frac{n + 1}{n} \max(X_1,X_2, ...,X_n) \)and\( \displaystyle Z_2 = \frac{2}{n} \sum_{i=1}^n X_i\)both have means equal to \(N\).Calculate the variance of \(Z_1\) and of \(Z_2\). Which estimator do you prefer, and why?

I don't intend to solve this since the only question of interest I can see is: "What is the pdf of \(Z_1\) (or the moment generating function if that is the approach you prefer) ?"

(For those interested I believe the origin of this question lies in the Korean war when the size of the production run of particular objects of interest was estimated from the serial numbers on recovered debris, it is also why it is common practice to assign quasi-random serial numbers to some types of military equipment)

(Another comment may be relevant, STEP is a pre-university exam, but the nature of many of the questions inclines me to post at least some of them in the University section of MHB)

CB

 

[jvicens]

Dear CB,

Thank you for bringing this interesting question to our attention. I am always intrigued by real-world applications of mathematical concepts.

To answer your question about the PDF of \(Z_1\) (or the moment generating function), let's start by breaking down the question and assumptions:

1. The hostile naval power possesses a large, unknown number \(N\) of submarines.
2. Interception of radio signals yields a small number \(n\) of their identification numbers \(X_i\).
3. The identification numbers are independent and uniformly distributed over the continuous range from \(0\) to \(N\).

Based on these assumptions, we can say that the probability of observing any particular identification number is \(\frac{1}{N}\). Now, let's define the maximum identification number observed as \(M = \max(X_1,X_2,...,X_n)\). The probability of observing a maximum identification number less than or equal to \(m\) is given by the cumulative distribution function (CDF) of the maximum of \(n\) independent and identically distributed (i.i.d) random variables:

\(F_M(m) = P(M \leq m) = \left(\frac{m}{N}\right)^n\)

Now, let's define \(Z_1\) as \(Z_1 = \frac{n+1}{n}M\). The PDF of \(Z_1\) can be obtained by differentiating the CDF with respect to \(m\):

\(f_{Z_1}(z_1) = \frac{dF_M(m)}{dm} = \frac{n(n+1)}{N^n}m^{n-1}\)

Now, to find the moment generating function (MGF) of \(Z_1\), we can use the fact that the MGF of a random variable \(Y\) is given by \(M_Y(t) = E[e^{tY}]\). Therefore, the MGF of \(Z_1\) can be calculated as:

\(M_{Z_1}(t) = E[e^{tZ_1}] = \int_{0}^{N} e^{tz_1} \frac{n(n+1)}{N^n}z_1^{n-1} dz_1\)

Using integration by parts, we can obtain the MGF of \(Z_1\) as:

\(M_{Z_1}(