Continuing from the previous problem, suppose X has c.d.f. given by F(x) = 1-e^(-sqrt(x)), x>0. a) Show that Y=sqrt(X) follows the exponential distribution with parameter 1. b) Using the previous result, compute the conditional probability P(X>=9|X>=4). The p.d.f. of an exponential randome variable with parameter λ is f(x) = λe^(-λx), for x>=0.
Question
Continuing from the previous problem, suppose X has c.d.f. given by F(x) = 1-e^(-sqrt(x)), x>0. a) Show that Y=sqrt(X) follows the exponential distribution with parameter 1. b) Using the previous result, compute the conditional probability P(X>=9|X>=4). The p.d.f. of an exponential randome variable with parameter λ is f(x) = λe^(-λx), for x>=0.
Solution
a) To show that Y=sqrt(X) follows the exponential distribution with parameter 1, we first need to find the cumulative distribution function (CDF) of Y.
The CDF of Y, denoted by F_Y(y), is given by P(Y<=y) = P(sqrt(X)<=y) = P(X<=y^2) = F_X(y^2).
Substituting F_X(x) = 1-e^(-sqrt(x)) into the equation, we get F_Y(y) = 1-e^(-y), for y>=0.
This is the CDF of an exponential distribution with parameter 1.
b) To compute the conditional probability P(X>=9|X>=4), we use the memoryless property of the exponential distribution, which states that P(X>=s+t|X>=s) = P(X>=t) for all s,t>=0.
Setting s=4 and t=5, we get P(X>=9|X>=4) = P(X>=5).
Since X follows the exponential distribution with parameter 1, we have P(X>=5) = e^(-5) = 0.0067.
Similar Questions
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Conditional distribution
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