GATE CS 2017 Set 2 Q.31 (CS Q.31): Probability and Statistics

For any discrete random variable $X$ , with probability mass function

P(X = j) = p_j, \quad p_j \ge 0, \quad j \in \{0, \dots, N\}, \quad \text{and} \quad \sum_{j=0}^N p_j = 1

define the polynomial function $g_X(z) = \sum_{j=0}^N p_j z^j$ . For a certain discrete random variable $Y$ , there exists a scalar $\beta \in [0, 1]$ such that $g_Y(z) = (1 - \beta + \beta z)^N$ . The expectation of $Y$ is

$N\beta(1 - \beta)$

$N\beta$

$N(1 - \beta)$

Not expressible in terms of $N$ and $\beta$ alone

Official answer and solution

Official answer: B
From: Official GATE 2017 answer key

Solution

Recall that the expectation of a discrete random variable can be obtained by taking the first derivative of its probability generating function evaluated at $z = 1$ .

For any discrete random variable $Y$ taking values in $\{0, 1, \dots, N\}$ with probability generating function $g_Y(z) = \sum_{j=0}^N P(Y = j) z^j$ :

g_Y'(z) = \sum_{j=0}^N j \cdot P(Y = j) z^{j-1}

Evaluating at $z = 1$ :

g_Y'(1) = \sum_{j=0}^N j \cdot P(Y = j) = E[Y]

Given $g_Y(z) = (1 - \beta + \beta z)^N$ , differentiate with respect to $z$ :

g_Y'(z) = N (1 - \beta + \beta z)^{N-1} \cdot \frac{d}{dz}(1 - \beta + \beta z) = N \beta (1 - \beta + \beta z)^{N-1}

Evaluate at $z = 1$ :

g_Y'(1) = N \beta (1 - \beta + \beta \cdot 1)^{N-1} = N \beta (1)^{N-1} = N\beta

Alternatively, notice that $g_Y(z) = (q + pz)^N$ is the exact probability generating function of a Binomial random variable with parameters $N$ and success probability $p = \beta$ , whose expected value is well-known to be $N\beta$ .

Final answer

Common trap

Confusing the expectation $E[Y] = N\beta$ with the variance $\text{Var}(Y) = N\beta(1 - \beta)$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 31 of 65, printed as CS Q.31
Section: Computer Science and Information Technology

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