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Official PYQ · 2016

GATE CS 2016 Set 1 Q04 — Probability: Random Variables and Distributions

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A probability density function on the interval [a,1][a, 1] is given by 1/x21/x^{2} and outside this interval the value of the function is zero. The value of aa is _________.
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Answer: 0.5

For a function f(x)f(x) to be a valid probability density function (PDF), the total area under the curve must be equal to 11:

\n\n∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x) \, dx = 1\n\n

Given that f(x)=1x2f(x) = \frac{1}{x^2} on the interval [a,1][a, 1] and 00 elsewhere, we can write:

\n\n∫a11x2 dx=1\int_{a}^{1} \frac{1}{x^2} \, dx = 1\n\n

Evaluating the definite integral:

\n\n[−1x]a1=1\left[ -\frac{1}{x} \right]_{a}^{1} = 1\n\n(−11)−(−1a)=1\left( -\frac{1}{1} \right) - \left( -\frac{1}{a} \right) = 1\n\n1a−1=1\frac{1}{a} - 1 = 1\n\n1a=2  ⟹  a=0.5\frac{1}{a} = 2 \implies a = 0.5\n\n

Thus, the value of aa is 0.5.

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