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 $2x_1{x}_2^3$, $P(\frac{1}{4}<X_1<\frac{1}{2}\mid X_2=\frac{5}{8})=0.4375$, and $P(\frac{1}{4}<X_1<\frac{1}{2})=0.1302$

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 $c_2$: $${\int }_0^1{c}_2{x}_2^{4}d{x}_2=1$$ Solving, we find that $c_2=5$. For $c_1$, we have $${\int }_0^{x_2}{c}_1\frac{x_1}{x_2^2}d{x}_1=1$$ which gives $c_1=2x_2$

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 $f_{1,2}(x_1,x_2)=f_{1\mid 2}(x_1\mid x_2)f_2(x_2)=2x_1{x}_2^3$

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 $$P(\frac{1}{4}<X_1<\frac{1}{2}\mid X_2=\frac{5}{8})={\int }_{1/4}^{1/2}{f}_{1\mid 2}(x_1\mid x_2)d{x}_1=0.4375$$

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 $$P(\frac{1}{4}<X_1<\frac{1}{2})={\int }_{1/4}^{1/2}{\int }_{x_1}^1{f}_{1,2}(x_1,x_2)d{x}_{2}d{x}_1=0.1302$$