Question

Let \(f_{1 \mid 2}\left(x_{1} \mid x_{2}\right)=c_{1} x_{1} / x_{2}^{2}, 0

Short answer

\(c_{1} = 2x_{2}\), \(c_{2} = 5\), the joint pdf of \(X_{1}\) and \(X_{2}\) is \(2x_{1}x_{2}^{3}\), \(P\left(\frac{1}{4}<X_{1}<\frac{1}{2} \mid X_{2}=\frac{5}{8}\right) = 0.4375\), and \(P\left(\frac{1}{4}<X_{1}<\frac{1}{2}\right) = 0.1302\)

Step-by-step solution

Determine the Constants \(c_{1}\) and \(c_{2}\)

For any pdf, the integral over all possible values must equal 1. Thus, we set up the following integrals to find \(c_{1}\) and \(c_{2}\): \[\int_{0}^{1} c_{2} x_{2}^4 dx_{2} = 1\] Solving, we find that \(c_{2} = 5\). For \(c_{1}\), we have \[\int_{0}^{x_{2}} c_{1} \frac{x_{1}}{x_{2}^{2}} dx_{1} = 1\] which gives \(c_{1} = 2 x_{2}\)

Compute the Joint pdf of \(X_{1}\) and \(X_{2}\)

The joint pdf of \(X_{1}\) and \(X_{2}\) can be obtained by multiplying the marginal pdf of \(X_{2}\) (which we have already found) with the conditional pdf of \(X_{1}\) (given \(X_{2} = x_{2}\)), so \(f_{1,2}(x_{1}, x_{2}) = f_{1 \mid 2}(x_{1} \mid x_{2}) f_{2}(x_{2})=2 x_{1} x_{2}^{3}\)

Calculate \(P\left(\frac{1}{4}

We want to calculate this specific probability. Plug \(x_{2} = \frac{5}{8}\) into the pdf \(f_{1 \mid 2}(x_{1} \mid x_{2})\), and then integrate from \(\frac{1}{4}\) to \(\frac{1}{2}\), so \[P\left(\frac{1}{4}<X_{1}<\frac{1}{2} \mid X_{2}=\frac{5}{8}\right)=\int_{1/4}^{1/2} f_{1 \mid 2}(x_{1} \mid x_{2}) dx_{1}= 0.4375\]

Calculate \(P\left(\frac{1}{4}

This probability can be calculated by integrating the joint pdf \(f_{1,2}(x_{1},x_{2})\) over the appropriate region. Here the region is a strip from \(\frac{1}{4}\) to \(\frac{1}{2}\) in the \(x_{1}\) direction, bounded by \(x_{2}=x_{1}\) and \(x_{2}=1\) in the \(x_{2}\) direction. So we obtain \[P\left(\frac{1}{4}<X_{1}<\frac{1}{2}\right)=\int_{1/4}^{1/2} \int_{x_{1}}^{1} f_{1,2}(x_{1},x_{2}) dx_{2} dx_{1} = 0.1302\]