A probability density function on the interval $[a, 1]$ is given by $1/x^2$ and outside this interval the value of the function is zero. The value of $a$ is ________ .

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Official answer: 0.5

Probability and Statistics

For $f(x)$ to be a valid probability density function, the total area under the curve must equal $1$ :

\int_{-\infty}^{\infty} f(x)\, dx = 1

Given $f(x) = \frac{1}{x^2}$ for $x \in [a, 1]$ and $0$ otherwise:

\int_{a}^{1} \frac{1}{x^2}\, dx = 1

Evaluating the integral:

\left[ -\frac{1}{x} \right]_{a}^{1} = 1

\left(-\frac{1}{1}\right) - \left(-\frac{1}{a}\right) = 1

-1 + \frac{1}{a} = 1

\frac{1}{a} = 2 \implies a = 0.5

Thus, the value of $a$ is 0.5.

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This Probability and Statistics previous year question is Q4 of the GATE CSE 2016 Set 1 paper. Find more GATE CSE PYQs on Probability and Statistics in Engineering Mathematics, or browse all GATE CSE PYQs.

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