Let X and Y be jointly distributed randomvariables such that the conditional distributionof Y, given X = x is uniform on the interval(x - 1, x 1) . Suppose E(X) = 1 andVar(X)=5/3 1.The mean of random variable Y is? 2.The variance of random variable Y is?

Question

Let X and Y be jointly distributed randomvariables such that the conditional distributionof Y, given X = x is uniform on the interval(x - 1, x   1) . Suppose E(X) = 1 andVar(X)=5/3 1.The mean of random variable Y is? 2.The variance of random variable Y is?

Answer

Mean and Variance of Random Variable Y

1. Mean of Random Variable Y

Given that the conditional distribution of Y, given X = x is uniform on the interval (x - 1, x + 1), we can find the mean of random variable Y.

The mean of random variable Y, denoted as E(Y), can be calculated using the law of total expectation:

E(Y) = E[E(Y|X)]

Since the conditional distribution of Y given X = x is uniform on the interval (x - 1, x + 1), the conditional mean of Y given X = x is the midpoint of this interval, which is x.

Therefore, E(Y|X) = x.

Taking the expectation of E(Y|X), we have:

E(Y) = E[X] = 1

So, the mean of random variable Y is 1.

2. Variance of Random Variable Y

To find the variance of random variable Y, denoted as Var(Y), we can use the law of total variance:

Var(Y) = E[Var(Y|X)] + Var[E(Y|X)]

First, let's find Var(Y|X). Since the conditional distribution of Y given X = x is uniform on the interval (x - 1, x + 1), the variance of a uniform distribution on the interval (a, b) is given by:

Var(Y|X) = (b - a)^2 / 12

In this case, a = x - 1 and b = x + 1. Substituting these values, we have:

Var(Y|X) = [(x + 1) - (x - 1)]^2 / 12 = 4/12 = 1/3

Next, let's find E(Y|X). As mentioned earlier, E(Y|X) = x.

Therefore, Var[E(Y|X)] = Var(X) = 5/3

Finally, substituting the values in the formula for the variance of Y:

Var(Y) = E[Var(Y|X)] + Var[E(Y|X)] = 1/3 + 5/3 = 6/3 = 2

So, the variance of random variable Y is 2.

Summary:
- The mean of random variable Y is 1.
- The variance of random variable Y is 2.

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