# Iterative expectation of continuous and discrete distributions

• cielo
In summary, the conditional distribution of Y given X is binomial (n, p=x), and the expected value of Y is found using the conditional distribution. To find the distribution of Y, use P[Y] = \int^{0}_{1} \left[nCy x^{y} (1-x)^{n-y} dx\right], which is a beta function.
cielo

## Homework Statement

Suppose X ~ uniform (0,1) and the conditional distribution of Y given X = x is binomial (n, p=x), i.e. P(Y=y|X=x) = nCy x$$^{y}$$ (1-x)$$^{n-y}$$ for y = 0, 1,..., n.

## Homework Equations

FInd E(y) and the distribution of Y.

## The Attempt at a Solution

f(x) = $$\frac{1}{b-a}$$ = $$\frac{1}{1-0}$$ =1E[Y] = E [E[Y|X=x]
= $$\int$$ E[Y|X=x] f(x) dx where the integral is from o to 1
= $$\int$$ [$$\Sigma$$ y f(y|x)] f(x) dx
= $$\int$$ [$$\Sigma$$ y nCy x$$^{y}$$ (1-x)$$^{n-y}$$] f(x) dx

cielo said:

## Homework Statement

Suppose X ~ uniform (0,1) and the conditional distribution of Y given X = x is binomial (n, p=x), i.e. P(Y=y|X=x) = nCy x$$^{y}$$ (1-x)$$^{n-y}$$ for y = 0, 1,..., n.

## Homework Equations

FInd E(y) and the distribution of Y.

## The Attempt at a Solution

f(x) = $$\frac{1}{b-a}$$ = $$\frac{1}{1-0}$$ =1

E[Y] = E [E[Y|X=x]
= $$\int$$ E[Y|X=x] f(x) dx where the integral is from o to 1
= $$\int$$ [$$\Sigma$$ y f(y|x)] f(x) dx
= $$\int$$ [$$\Sigma$$ y nCy x$$^{y}$$ (1-x)$$^{n-y}$$] f(x) dx

You know the conditional distribution of Y given X. Use that to find E[Y | X]. The answer is a function of X - find its expectation with respect to X to get E[E[Y |X]] = E[Y]

Thank you so much for your very good idea. Because of that, I already got the E[Y].

Can you still help me in finding the distribution of Y?

P[Y] = $$\int^{0}_{1}$$ $$\left[nCy x^{y} (1-x)^{n-y} dx\right]$$

I understand that is a a beta function if we ignore the constant. But can you help me find the final distribution of Y?

## What is the difference between continuous and discrete distributions?

Continuous distributions describe variables that can take on any value within a certain range, while discrete distributions describe variables that can only take on a specific set of values.

## What is the iterative expectation method used for?

The iterative expectation method is used to calculate the expected value of a distribution by breaking it down into smaller, simpler parts and then summing them together.

## How do you calculate the iterative expectation for a continuous distribution?

To calculate the iterative expectation for a continuous distribution, you first need to find the probability density function (PDF) and the expected value for each subinterval. Then, you multiply each expected value by its corresponding probability and sum them together to get the overall expected value.

## Can the iterative expectation method be used for any type of distribution?

Yes, the iterative expectation method can be used for any type of distribution, whether it is continuous or discrete. It is a general method for calculating expected values and can be applied to various distributions.

## What are the advantages of using the iterative expectation method?

One advantage of using the iterative expectation method is that it allows for complex distributions to be broken down into smaller, more manageable parts. It also provides a more accurate estimate of the expected value compared to other simpler methods, such as the mean or median.

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