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Let $X_1$ and $X_2$ be two independent exponentially distributed random variables with means $0.5$ and $0.25$, respectively. Then $Y=\text{min}(X_1, X_2)$ is

- exponentially distributed with mean $1/6$
- exponentially distributed with mean $2$
- normally distributed with mean $3/4$
- normally distributed with mean $1/6$

1 vote

Best answer

Mean of X1 = 0.5

Mean of X2 = 0.25

Parameter for X1 = 1/0.5 = 2

Parameter for X2 = 1/0.25 = 4

Since, Y = min(X_{1},X2), Y is exponentially Distributed and it's Parameter = Parameter 1 + Parameter 2 = 2+4 = 6

(PROOF : http://www.math.wm.edu/~leemis/chart/UDR/PDFs/ExponentialM.pdf)

Mean of Y = 1/Parameter = 1/6

**Hence, A is Correct!**