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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
- a)exponentially distributed with mean 1⁄6
- b)exponentially distributed with mean 2
- c)normally distributed with mean 3⁄4
- d)normally distributed with mean 1⁄6
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Let X1 and X2 be two independent exponentially distributed random vari...
To find the distribution of Y, we can use the fact that P(Y > y) = P(X1 > y and X2 > y) since Y is the minimum of X1 and X2.
Since X1 and X2 are independent, we can use the fact that the probability of the intersection of two events is the product of their probabilities: P(X1 > y and X2 > y) = P(X1 > y) * P(X2 > y).
Using the exponential distribution formula, we have P(X1 > y) = e^(-y/0.5) and P(X2 > y) = e^(-y/0.25). Therefore,
P(Y > y) = P(X1 > y and X2 > y) = P(X1 > y) * P(X2 > y) = e^(-y/0.5) * e^(-y/0.25) = e^(-3y/4).
This is the cumulative distribution function (CDF) of Y. To find the probability density function (PDF), we can take the derivative of the CDF:
f(y) = d/dy (e^(-3y/4)) = (-3/4) * e^(-3y/4).
This is the PDF of Y, which is an exponential distribution with mean 4/3. Therefore, the statement that Y is exponentially distributed with mean 1 is false.
Since X1 and X2 are independent, we can use the fact that the probability of the intersection of two events is the product of their probabilities: P(X1 > y and X2 > y) = P(X1 > y) * P(X2 > y).
Using the exponential distribution formula, we have P(X1 > y) = e^(-y/0.5) and P(X2 > y) = e^(-y/0.25). Therefore,
P(Y > y) = P(X1 > y and X2 > y) = P(X1 > y) * P(X2 > y) = e^(-y/0.5) * e^(-y/0.25) = e^(-3y/4).
This is the cumulative distribution function (CDF) of Y. To find the probability density function (PDF), we can take the derivative of the CDF:
f(y) = d/dy (e^(-3y/4)) = (-3/4) * e^(-3y/4).
This is the PDF of Y, which is an exponential distribution with mean 4/3. Therefore, the statement that Y is exponentially distributed with mean 1 is false.
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Let X1 and X2 be two independent exponentially distributed random variables with means 0.5 and 0.25, respectively. Then Y = min (X1, X2) isa)exponentially distributed with mean 1⁄6b)exponentially distributed with mean 2c)normally distributed with mean 3⁄4d)normally distributed with mean 1⁄6Correct answer is option 'A'. Can you explain this answer?
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