Project ForgeEngineering Mathematics
Calculus
GATE 2018
Medium
Let X1 and X2 be two independent exponentially distributed random variables with means 0.5 and 0.25, respectively. Then Y = min(X1, X2) is
- exponentially distributed with mean 1/6
- Bexponentially distributed with mean 2
- Cnormally distributed with mean 3/4
- Dnormally distributed with mean 1/6
Explanation
The rates are λ1 = 1/0.5 = 2 and λ2 = 1/0.25 = 4. The minimum of independent exponential random variables is itself exponential with rate equal to the sum of the individual rates: λ = λ1+λ2 = 6, so Y is exponential with mean 1/6.
Source: GATE ME 2018 Session 2 (Set 2) (2018)
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