Question

LetXbe a random variable with a continuous distribution.

Let \({{\bf{x}}_{\bf{1}}}{\bf{ = }}{{\bf{x}}_{\bf{2}}}{\bf{ = x}}\)

a.Prove that both \({{\bf{x}}_{\bf{1}}}\) and \({{\bf{x}}_{\bf{2}}}\) have a continuous distribution.

b.Prove that \({\bf{x = (}}{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{)}}\)does not have a continuous

joint distribution.

Short answer

1. \({x_1}\)and\({x_2}\)both have continuous distribution.
2. \({x_1}\)and \({x_2}\)does not have joint continuous distribution.

Step-by-step solution

Given Information

Here given distributions are continuous distributions. We have to show the distributions are continuous and do not have a continuous joint distribution.

State the distributions

Here \(x\)have continuous distribution. And \({x_1} = {x_2} = x\) .

Let for this problem we assume that \(x\)follows uniform distribution. Which is written as \(x \sim U\left( {a,b} \right)\) .

Prove that \({x_1}\) and \({x_2}\) have a continuous distribution

The pdf of \({x_1}\) is given by

\(\begin{align}f\left( {{x_1}} \right) = \int_a^b {\frac{1}{{b - a}}dx} \\\end{align}\)

We know that sum of all pdf is\(1\). Then the pdf of\({x_1}\)is given by

\(\begin{align}f\left( {{x_1}} \right) &= \int_a^b {\frac{1}{{b - a}}dx} \\ &= \frac{{b - a}}{{b - a}}\\ &= 1\end{align}\)

Similarly for the pdf of\({x_2}\)is given by

\(\begin{align}f\left( {{x_2}} \right) &= \int_a^b {\frac{1}{{b - a}}dx} \\ &= \frac{{b - a}}{{b - a}}\\ &= 1\end{align}\)

Hence here we have to show \({x_1}\)and \({x_2}\) both have continuous distribution.

Prove that \({\bf{x = }}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}} \right)\) does not have a continuous distribution

If we multiply the two pdf of \({x_1}\)and \({x_2}\) then it is given by

\(\begin{align}f\left( x \right) &= f\left( {{x_1}} \right) \times f\left( {{x_2}} \right)\\ &= \int_a^b {\frac{1}{{b - a}}dx \times \int_a^b {\frac{1}{{b - a}}} } dx\\ &= \int_a^b {\frac{1}{{{{\left( {b - a} \right)}^2}}}dx} \\ &= \frac{1}{{\left( {b - a} \right)}}\end{align}\)

Hence the pdf of \({x_1}\)and \({x_2}\) does not have joint continuous distribution.