Exam-Style Problem

9231 P41 - Nov 2025 - Q7 - 10 marks

6607

A discrete random variable \(X\) takes values \(r=0,1,2\) with probabilities \(\mathrm{P}(X=r)\) as given in the following table.

\(r\)	\(0\)	\(1\)	\(2\)
\(\mathrm{P}(X=r)\)	\(a\)	\(2a\)	\(b\)

(a) Write down the probability generating function of \(X\), and use it to find an expression for \(\mathrm{E}(X)\) in terms of \(a\) and \(b\).

(b) Show that \(\operatorname{Var}(X)=2b+2(a+b)(1-2a-2b)\).

The random variable \(Y\) is defined by \(Y=X_1+X_2+X_3+\cdots+X_{10}\), where \(X_1,X_2,X_3,\ldots,X_{10}\) are ten independent observations of \(X\).

(c) Using the probability generating function of \(Y\), and your answer to part (a), show that \(\mathrm{E}(Y)=10\mathrm{E}(X)\).

(d) For the case \(b=0\), define fully the distribution of \(Y\).

Solution

Answer:

(a) \(G_X(t)=a+2at+bt^2\), and \(\mathrm{E}(X)=2a+2b\).
(b) \(\operatorname{Var}(X)=2b+2(a+b)(1-2a-2b)\).
(c) \(\mathrm{E}(Y)=10\mathrm{E}(X)\).
(d) When \(b=0\), \(Y\sim \mathrm{B}\left(10,\frac{2}{3}\right)\). Equivalently, \(\mathrm{P}(Y=y)=\binom{10}{y}\left(\frac{2}{3}\right)^y\left(\frac{1}{3}\right)^{10-y}\), for \(y=0,1,\ldots,10\).

(a) The probability generating function is \(G_X(t)=\mathrm{E}(t^X)\). Hence
\(G_X(t)=a t^0+2a t^1+b t^2=a+2at+bt^2\).
For a probability generating function, \(\mathrm{E}(X)=G_X'(1)\). Differentiating gives
\(G_X'(t)=2a+2bt\),
so
\(\mathrm{E}(X)=G_X'(1)=2a+2b\).
(b) Using the probability generating function formula
\(\operatorname{Var}(X)=G_X''(1)+G_X'(1)-\{G_X'(1)\}^2\).
From \(G_X'(t)=2a+2bt\),
\(G_X''(t)=2b\), so \(G_X''(1)=2b\).
Therefore
\(\operatorname{Var}(X)=2b+(2a+2b)-(2a+2b)^2\).
Factorising the last two terms,
\(\operatorname{Var}(X)=2b+(2a+2b)\{1-(2a+2b)\}\).
Since \(2a+2b=2(a+b)\), this becomes
\(\operatorname{Var}(X)=2b+2(a+b)(1-2a-2b)\), as required.
(c) Since \(Y=X_1+X_2+\cdots+X_{10}\), where the \(X_i\) are independent and each has the same distribution as \(X\), the probability generating function of \(Y\) is
\(G_Y(t)=\{G_X(t)\}^{10}=(a+2at+bt^2)^{10}\).
Differentiating,
\(G_Y'(t)=10(a+2at+bt^2)^9(2a+2bt)\).
So
\(G_Y'(t)=20(a+bt)(a+2at+bt^2)^9\).
Hence
\(\mathrm{E}(Y)=G_Y'(1)=20(a+b)(a+2a+b)^9=20(a+b)(3a+b)^9\).
The probabilities for \(X\) must sum to 1, so
\(a+2a+b=1\), giving \(3a+b=1\).
Thus
\(\mathrm{E}(Y)=20(a+b)\).
From part (a), \(\mathrm{E}(X)=2a+2b=2(a+b)\), so
\(10\mathrm{E}(X)=10\times 2(a+b)=20(a+b)\).
Therefore
\(\mathrm{E}(Y)=10\mathrm{E}(X)\).
(d) If \(b=0\), then the probabilities sum to 1 gives
\(3a+0=1\), so \(a=\frac{1}{3}\).
Thus
\(\mathrm{P}(X=0)=\frac{1}{3}\), \(\mathrm{P}(X=1)=\frac{2}{3}\), and \(\mathrm{P}(X=2)=0\).
So each \(X_i\) is a Bernoulli random variable with probability \(\frac{2}{3}\) of taking the value 1. Therefore the sum of ten independent such variables has a binomial distribution:
\(Y\sim \mathrm{B}\left(10,\frac{2}{3}\right)\).
Fully, this means
\(\mathrm{P}(Y=y)=\binom{10}{y}\left(\frac{2}{3}\right)^y\left(\frac{1}{3}\right)^{10-y}\), for \(y=0,1,\ldots,10\).