Question

Let \(f(x)\) and \(F(x)\) denote, respectively, the pdf and the cdf of the random variable \(X\). The conditional pdf of \(X\), given \(X>x_{0}, x_{0}\) a fixed number, is defined by \(f\left(x \mid X>x_{0}\right)=f(x) /\left[1-F\left(x_{0}\right)\right], x_{0}x_{0}\right)\) is a pdf. (b) Let \(f(x)=e^{-x}, 02 \mid X>1)\).

Short answer

Part (a): The function \(f\left(x \mid X>x_{0}\right)\) is a valid pdf as it is non-negative for all \(x>\,x_{0}\) and the total integral over its entire range equals 1 due to normalization constant \([1-F\left(x_{0}\right)]\). Part (b): To compute the conditional probability \(P(X>2 \mid X>1)\), we can use properties of the exponential distribution, yielding a result of \(e^{-1}\).

Step-by-step solution

Validation of conditional pdf

To show that \(f\left(x \mid X>x_{0}\right)\) is a pdf, it has to meet two conditions: it should be non-negative for all \(x>\ x_{0}\), and the total probability must equal 1 when integrated over the entire range (from \(x_{0}\) to \(\infty\)). So we express the conditional pdf as \(f\left(x \mid X>x_{0}\right)=f(x) /\left[1-F\left(x_{0}\right)\right]\) for \(x_{0}<x\), zero elsewhere and replace \(f(x)\) with general pdf function, ensuring these conditions are met.

Calculate the normalization constant

To ensure the total integral equals 1, we need to determine a normalization constant. This is given by \[1-F\left(x_{0}\right)\], which is the probability that \(X\) takes a value greater than the specific value \(x_{0}\). Therefore, the denominator acts as a normalization constant for the conditional probability that ensures the total probability equals 1.

Computation of conditional probability

We are given the exponential pdf function \(f(x)\)=\(e^{-x}\) for \(02 \mid X>1)\). Using the definition of the conditional probability and the exponential distribution properties, we compute this probability.