Question

3. (15 marks) Conditional on X,Y is a Bern(X) random variable. Marginally, X is a Beta(α,β) random variable where both α>0 and β>0 are constants. Note that E[Y∣X]=X and Var(Y∣X)=X(1−X) (a) Determine the marginal probability mass function (PMF) of Y. [5 marks] (b) Determine the conditional probability density function (PDF) of X given Y. [5 marks] (c) Conditional on X,(Y1​,…,Yn​) are independent and identically distributed from a Bern(X) distribution. Again X is a Beta(α,β) random variable where both α>0 and β>0 are constants. Determine the conditional probability density function (PDF) of X given Y1​,…,Yn​ [5 marks]

Answer 1

(a) Marginal Probability Mass Function (PMF) of Y:

`\[ P(Y=y) = (\Gamma(\alpha+\beta))/(\Gamma(\alpha)\Gamma(\beta)) \cdot y^(\alpha-1) \cdot (1-y)^(\beta-1) \]`

(b) Conditional Probability Density Function (PDF) of X given Y:

`\[ f_(X|Y)(x|y) = (1)/(B(\alpha,\beta)) \cdot x^(\alpha y - 1) \cdot (1-x)^(\beta(1-y) - 1) \]`

(c) Conditional Probability Density Function (PDF) of X given
`\(Y_1, \ldots, Y_n\):`

`\[ f_(X|Y_1,\ldots,Y_n)(x|y_1,\ldots,y_n) = \frac{1}{B(\alpha+n\bar{y},\beta+n(1-\bar{y}))} \cdot x^{\alpha+n\bar{y}-1} \cdot (1-x)^{\beta+n(1-\bar{y})-1} \]`

Step-by-step explanation:

(a) Marginal Probability Mass Function (PMF) of Y:

The marginal PMF of Y is derived from the integration of the conditional distribution
`\(Y|X\)`, which is a Bernoulli distribution with probability X, over the marginal distribution of X, which is a Beta distribution with parameters
`\(\alpha\) and \(\beta\)`. The formula for the marginal PMF of Y is given by:

`\[ P(Y=y) = (\Gamma(\alpha+\beta))/(\Gamma(\alpha)\Gamma(\beta)) \cdot y^(\alpha-1) \cdot (1-y)^(\beta-1) \]`

Here,
`\(\Gamma\)` represents the gamma function, and the expression combines the probabilities from the Beta distribution with the probabilities of the Bernoulli distribution.

(b) Conditional Probability Density Function (PDF) of X given Y:

To find the conditional PDF of X given Y, we use Bayes' theorem. Given that
`\(E[Y|X]=X\) and \(Var(Y|X)=X(1-X)\)`, the conditional distribution is a Bernoulli distribution. Applying Bayes' theorem, the conditional PDF is expressed as:

`\[ f_(X|Y)(x|y) = (1)/(B(\alpha,\beta)) \cdot x^(\alpha y - 1) \cdot (1-x)^(\beta(1-y) - 1) \]`

This formula incorporates the information about the mean and variance of Y given X and utilizes the Beta function
`\(B(\alpha,\beta)\)` to normalize the distribution.

(c) Conditional Probability Density Function (PDF) of X given
`\(Y_1, \ldots, Y_n\):`

Extending the conditional distribution to n independent and identically distributed Bernoulli random variables, the conditional PDF of X given
`\(Y_1, \ldots, Y_n\)` is obtained. The parameters of the Beta distribution are updated based on the sample mean
`\(\bar{y}\)` and the number of observations n. The formula is given by:

`\[ f_(X|Y_1,\ldots,Y_n)(x|y_1,\ldots,y_n) = \frac{1}{B(\alpha+n\bar{y},\beta+n(1-\bar{y}))} \cdot x^{\alpha+n\bar{y}-1} \cdot (1-x)^{\beta+n(1-\bar{y})-1} \]`

This expression reflects the cumulative information from the observed Bernoulli variables, providing a updated probability distribution for X given the observed values of
`\(Y_1, \ldots, Y_n\).`