GATE 2018 | MATHS | Q-62

Question

Let $X1, X2, X3, X4$ be independent exponential random variables with mean $1, 1/2, 1/3, 1/4,$ respectively. Then $Y = min(X1, X2, X3, X4)$ has exponential distribution with mean equal to

Answer 1

To find the mean of the minimum of independent exponential random variables, we can use the fact that the minimum of exponential random variables follows an exponential distribution with a rate equal to the sum of the rates of the individual exponentials.

Let \(X_i\) denote the exponential random variable with mean \(1/i\) for \(i = 1, 2, 3, 4\). The rate parameter \(\lambda_i\) for each \(X_i\) is given by \(\lambda_i = 1/(1/i) = i\).

The minimum \(Y = \min(X_1, X_2, X_3, X_4)\) has a rate parameter \(\lambda_Y\) equal to the sum of the individual rates:
\[
\lambda_Y = \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 = 1 + 2 + 3 + 4 = 10.
\]

Therefore, \(Y\) follows an exponential distribution with mean \(1/\lambda_Y\):
\[
\text{Mean of } Y = \frac{1}{\lambda_Y} = \frac{1}{10}.
\]

So, the mean of \(Y\) is \(\frac{1}{10}\).