GATE 2023 IN Q.32: solution and answer

GATE 2023 · INQ.32 · 1 mark · NAT

$X$ is a discrete random variable which takes values 0, 1 and 2. The probabilities are $P (X = 0) = 0.25$ and $P (X = 1) = 0.5$ . With $E [.]$ denoting the expectation operator, the value of $E [X] - E [X^{2}]$ is ______ (rounded off to one decimal place).

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Answer and solution

ANSWER

$E [X] - E [X^{2}] = - 0.5$

Official answer: -0.5Source: Official answer key, GATE 2023 IN (IIT Kanpur)

The probabilities must add up to 1, so $P (X = 2) = 1 - 0.25 - 0.5 = 0.25$ .
Mean, from $E [X] = \sum x P (X = x)$ : $E [X] = 0 (0.25) + 1 (0.5) + 2 (0.25) = 0 + 0.5 + 0.5 = 1$ .
Second moment, from $E [g (X)] = \sum g (x) P (X = x)$ with $g (x) = x^{2}$ : $E [X^{2}] = 0^{2} (0.25) + 1^{2} (0.5) + 2^{2} (0.25) = 0 + 0.5 + 1 = 1.5$ .
Therefore $E [X] - E [X^{2}] = 1 - 1.5 = - 0.5$
Check: by linearity, $E [X] - E [X^{2}] = E [X (1 - X)]$ . The product $x (1 - x)$ is $0$ at $x = 0$ and at $x = 1$ , and $2 (1 - 2) = - 2$ at $x = 2$ . So the value is $- 2 \times 0.25 = - 0.5$ , which agrees.

KEY IDEA

First find the missing probability from $\sum P = 1$ . Then use $E [g (X)] = \sum g (x) P (X = x)$ : take $g (x) = x$ for the mean and $g (x) = x^{2}$ for the second moment. Linearity also gives the answer in one line as $E [X (1 - X)]$ , and only $x = 2$ adds anything to that sum.

WATCH OUT FOR

There are two common slips. One is leaving out the $x = 2$ term because its probability is not printed, when it must be found as $1 - 0.25 - 0.5 = 0.25$ . The other is computing $E [X^{2}]$ as $(E [X])^{2} = 1$ , which gives $0$ instead of $- 0.5$ . $E [X^{2}]$ is the second moment, not the square of the mean. The two differ by the variance, which here is $0.5$ .

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GATE 2023 IN Q.32: solution and answer

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