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Random vectors - Exercise set 2

This exercise set contains some solved exercises on absolutely continuous random vectors and joint probability density functions. The theory needed to solve these exercises is introduced in the lecture entitled Random vectors.

Exercise 2.1

Let X be a $2 imes 1$ absolutely continuous random vector and denote its components by X_1 and X_2. Let the support of X be [eq1] i.e. the set of all $2 imes 1$ vectors such that the first component belongs to the interval $left[ 0,2 ight] $ and the second component belongs to the interval $left[ 0,3 ight] $. Let the joint probability density function of X be:[eq2]Compute [eq3].

nav_button Solution

By the very definition of joint probability density function:[eq4]

Exercise 2.2

Let X be a $2 imes 1$ absolutely continuous random vector and denote its components by X_1 and X_2. Let the support of X be [eq5]i.e. the set of all $2 imes 1$ vectors such that the first component belongs to the interval [eq6] and the second component belongs to the interval $left[ 0,2 ight] $. Let the joint probability density function of X be:[eq7]Compute [eq8].

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First of all note that $X_{1}+X_{2}leq 3$ if and only if $X_{2}leq 3-X_{1}$. Using the definition of joint probability density function, we obtain:[eq9]

Now, note that, when [eq10], the inner integral is[eq11]Therefore:[eq12]

Exercise 2.3

Let X be a $2 imes 1$ absolutely continuous random vector and denote its components by X_1 and X_2. Let the support of X be [eq13] (i.e. the set of all $2$-dimensional vectors with positive entries) and its joint probability density function be:[eq14]Derive the marginal probability density functions of X_1 and X_2.

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The support of X_1 is:[eq15](recall that [eq16] and [eq17])

We can find the marginal density by integrating the joint density with respect to $x_{2}$:[eq18]




When $x<0$, then [eq19] and the above integral is trivially equal to 0. Thus, when $x<0$, then [eq20].

When $x>0$, then:[eq21]but the first of the two integrals is zero since [eq22] when $x_{2}<0$; as a consequence:[eq23]So, putting pieces together, the marginal density function of X_1 is:[eq24]

Obviously, by symmetry, the marginal density function of X_2 is:[eq25]

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