Question

Let (𝑋, 𝑌, 𝑍) be a random vector having the joint probability density function
\(f(x,y, z) =\begin{cases} \frac{1}{2\,xy}, & \quad \text{if }0<z<y<x<1.\\ \frac{1}{2x^2}, & \quad if\,0<z<x<y<2x<2,\\0, & \quad\,\,Otherwise.\end{cases}\)
Then, which one of the following statements is FALSE?

• A. \(𝑃(𝑍 < 𝑌 < 𝑋) = \frac{1}{2}\)
• B. 𝑃(𝑋 < 𝑌 < 𝑍) = 0
• C. 𝐸(min{𝑋, 𝑌}) =\(\frac{1}{4}\)
• D. 𝑉𝑎𝑟 (𝑌 | 𝑋 = \(\frac{1}{2}\) ) = \(\frac{1}{12}\)

Answer 1

The Correct Option is C

This question requires us to analyze the joint probability density function (pdf) of a random vector \( (X, Y, Z) \) and determine which provided statement is false. We'll evaluate each statement based on the given conditions of the pdf.

Joint pdf and Conditions: 

The joint pdf is given as:

\[f(x, y, z) = \begin{cases} \frac{1}{2xy}, & \text{if } 0 < z < y < x < 1\\ \frac{1}{2x^2}, & \text{if } 0 < z < x < y < 2x < 2\\ 0, & \text{Otherwise} \end{cases}\]

Analysis of Options:

1. \(P(Z < Y < X) = \frac{1}{2}\)

To find this probability, we need to integrate the pdf over the region defined by \(0 < z < y < x < 1\). This can be shown by solving the triple integral:

\[\int_0^1 \int_0^x \int_0^y \frac{1}{2xy} \, dz \, dy \, dx\]

This evaluates to \(\frac{1}{2}\), confirming the statement is true.

1. \(P(X < Y < Z) = 0\)

According to the conditions \(0 < z < x < y < 2x < 2\), there is no valid region for \(X < Y < Z\), therefore this probability is indeed 0, verifying the statement is true.

1. \(E(\min\{X, Y\}) = \frac{1}{4}\)

The expectation, \(E(\min\{X, Y\})\) involves integration over the respective region and computing \( \min\{x, y\} \). Based on settings \(0 < z < y < x < 1\), this requires complex integration, which shows that the calculated expectation is not \(\frac{1}{4}\) but rather a different value. Therefore, this statement is false.

1. \(Var(Y | X = \frac{1}{2}) = \frac{1}{12}\)

For this, compute the conditional variance \(Var(Y | X = \frac{1}{2})\) by integrating over the possible values of \(Y\) given \(X = \frac{1}{2}\). Conditional pdf can be deduced using:

\[f(y|x) \propto \int f(x, y, z) \, dz\]

This involves specific conditions, and computing further shows that the variance is indeed \(\frac{1}{12}\), hence the statement is true.

Conclusion:

Based on the analyses, the statement "\(E(\min\{X, Y\}) = \frac{1}{4}\)" is false. Thus, it is the correct answer to the question regarding which statement is FALSE.