Answer and Step-by-step explanation:
To answer the question about X(1) and X(2) as exponential random variables:
(a) X(1) = min{X1, X2} is indeed an exponential random variable. In order to show this, we need to demonstrate that X(1) follows the properties of an exponential distribution.
The exponential distribution is characterized by its probability density function (PDF), given by f(x) = (1/θ)e^(-x/θ), where θ is the parameter representing the mean or average of the distribution.
Since X1 and X2 are independent exponential random variables with the same parameter θ, their PDFs are both (1/θ)e^(-x/θ).
To find the PDF of X(1), we consider the event X(1) > x, where x is a given value. This event occurs when both X1 and X2 are greater than x.
The probability of X1 > x is given by P(X1 > x) = ∫[x, ∞] (1/θ)e^(-t/θ) dt, and the probability of X2 > x is given by P(X2 > x) = ∫[x, ∞] (1/θ)e^(-t/θ) dt.
Since X1 and X2 are independent, the probability of both X1 > x and X2 > x is obtained by multiplying their individual probabilities:
P(X(1) > x) = P(X1 > x) * P(X2 > x) = ∫[x, ∞] (1/θ)e^(-t/θ) dt * ∫[x, ∞] (1/θ)e^(-t/θ) dt.
Simplifying this expression, we have P(X(1) > x) = (e^(-x/θ))^2 = e^(-2x/θ).
The complement of this probability represents the cumulative distribution function (CDF) of X(1). Taking the derivative of the CDF with respect to x gives us the PDF of X(1):
f(x) = d/dx(1 - e^(-2x/θ)) = (2/θ)e^(-2x/θ).
Comparing this PDF with the standard form of the exponential distribution, we find that X(1) is indeed an exponential random variable with parameter 2θ.
(b) On the other hand, X(2) = max{X1, X2} is not an exponential random variable. The maximum of two exponential random variables does not follow the exponential distribution.
To find the PDF of X(2), we need to consider the event X(2) > x, where x is a given value. This event occurs when at least one of X1 or X2 is greater than x.
The probability of X1 > x is given by P(X1 > x) = ∫[x, ∞] (1/θ)e^(-t/θ) dt, and the probability of X2 > x is given by P(X2 > x) = ∫[x, ∞] (1/θ)e^(-t/θ) dt.
However, calculating the probability of either X1 or X2 being greater than x is not sufficient to find the PDF of X(2). We need to consider all possible combinations of X1 and X2 that satisfy X(2) > x.
This involves finding the joint probability distribution of X1 and X2 and determining the region where X(2) > x. The resulting PDF of X(2) will depend on the specific distribution of X1 and X2, and it will not follow the exponential distribution.