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Conditional probability distributions - Exercise set 1

This exercise set contains some solved exercises on conditional probability distributions. The theory needed to solve these exercises is introduced in the lecture entitled Conditional probability distributions.

Exercise 1.1

Let [eq1] be a discrete random vector with support: [eq2]and joint probability mass function:[eq3]Compute the conditional probability mass function of X given $Y=0$.

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The marginal probability mass function of Y evaluated at $y=0$ is:[eq4]The support of X is:[eq5]Thus, the conditional probability mass function of X given $Y=0$ is:[eq6]

Exercise 1.2

Let [eq1] be an absolutely continuous random vector with support: [eq8]and its joint probability density function be:[eq9]Compute the conditional probability density function of X given $Y=2$.

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The support of Y is:[eq10]When [eq11], the marginal probability density function of Y is [eq12]; when [eq13], the marginal probability density function of Y is:[eq14]Thus, the marginal probability density function of Y is:[eq15]When evaluated at the point $y=2$, it becomes:[eq16]The support of X is:[eq17]Thus, the conditional probability density function of X given $Y=2$ is:[eq18]

Exercise 1.3

Let X be an absolutely continuous random variable with support[eq19]and probability density function[eq20]Let Y be another absolutely continuous random variable with support[eq21]and conditional probability density function[eq22]Find the marginal probability density function function of Y.

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The support of the vector [eq1] is:[eq24]and the joint probability function of X and Y is:[eq25]The marginal probability density function of Y is obtained by marginalization, integrating x out of the joint probability density function:

[eq26]Thus, for [eq27] we trivially have [eq28] (because [eq29]), while for [eq30] we have:[eq31]Thus, the marginal probability density function of Y is:[eq32]

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