Understanding Independence of Random Variables in the Discrete Case

 Show in the discrete case that if X and Y are independent, then E[X|Y = y] = E[X] for all y.

In probability and statistics, we often deal with random variables, which are variables that take on a set of possible values with a certain probability. Independence between random variables X and Y is a property that states that the occurrence of one event has no effect on the occurrence of the other. In this blog post, we will demonstrate how independence between two discrete random variables X and Y leads to the result that E[X|Y=y] = E[X] for all y.

Expected Value of Independent Random Variables:

The expected value of a random variable X is denoted as E[X] and represents the average or expected outcome of a random event. It is calculated as the sum of all possible values of X multiplied by their respective probabilities. For example, if X takes on the values {1,2,3} with probabilities {0.1,0.5,0.4}, then E[X] can be calculated as:

E[X] = 1*0.1 + 2*0.5 + 3*0.4 = 2.3

Conditional Expectation:

The conditional expectation of a random variable X given an event Y is denoted as E[X|Y=y] and represents the expected value of X given that event Y has occurred. It is calculated as the sum of all possible values of X multiplied by their respective probabilities, conditioned on Y=y. For example, if X takes on the values {1,2,3} with probabilities {0.1,0.5,0.4}, and Y=1, then E[X|Y=1] can be calculated as:

E[X|Y=1] = 1*0.1 + 2*0.5 + 3*0.4 = 2.3

Independence and Conditional Expectation:

Now, let's consider the case where X and Y are independent random variables. Independence between X and Y means that the occurrence of one event has no effect on the occurrence of the other. In other words, the probability of X given Y=y is the same as the probability of X. This means that:

P(X=x|Y=y) = P(X=x) for all x,y

Using this property, we can show that E[X|Y=y] = E[X] for all y. To see this, consider the following calculation:

E[X|Y=y] = Sum over x of x * P(X=x|Y=y)

= Sum over x of x * P(X=x) (since X and Y are independent)

= E[X]

Therefore, we have shown that if X and Y are independent, then E[X|Y=y] = E[X] for all y in the discrete case.

In this blog post, we have demonstrated how independence between two discrete random variables X and Y leads to the result that E[X|Y=y] = E[X] for all y. This result is important in understanding the relationship between independence and conditional expectations and can be applied in various statistical and probabilistic settings.

Calculating the Expectations of Successive Dice Rolls

 An unbiased die is successively rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a six and a five. 

Find (a) E[X],

(b) E[X|Y = 1]

In this blog post, we will discuss the expectations of the number of rolls necessary to obtain a six (X) and a five (Y) in a sequence of successive rolls of an unbiased die. We will calculate (a) E[X], (b) E[X|Y=1], and (c) E[X|Y=5].

Calculating E[X]:

The expected value of X is the average number of rolls necessary to obtain a six. Let's assume that the probability of obtaining a six in one roll is 1/6. Therefore, the probability of not obtaining a six in one roll is 5/6. To calculate E[X], we can use the formula:

E[X] = 1/P(X=1) + 2/P(X=2) + 3/P(X=3) + ...

Where P(X=n) is the probability of obtaining a six in n rolls.

P(X=1) = 1/6, P(X=2) = (5/6) * (1/6), P(X=3) = (5/6)^2 * (1/6), ...

Therefore, the expected value of X is:

E[X] = 1/1/6 + 2/(5/6) * (1/6) + 3/(5/6)^2 * (1/6) + ...

E[X] = 6.

Calculating E[X|Y=1]:

The expected value of X given Y=1 is the average number of rolls necessary to obtain a six, given that a five was obtained in the first roll. To calculate E[X|Y=1], we use the formula:

E[X|Y=1] = 1/P(X=1|Y=1) + 2/P(X=2|Y=1) + 3/P(X=3|Y=1) + ...

Where P(X=n|Y=1) is the probability of obtaining a six in n rolls given that a five was obtained in the first roll.

P(X=1|Y=1) = 0, P(X=2|Y=1) = 1/6, P(X=3|Y=1) = (5/6) * (1/6), ...

Therefore, the expected value of X given Y=1 is:

E[X|Y=1] = 1/0 + 2/1/6 + 3/(5/6) * (1/6) + ...

E[X|Y=1] = 7.

Solving Conditional Expectations of Joint Probabilities with X, Y, Z

Suppose p(x, y, z), the joint probability mass function of the random variables X,

Y, and Z, is given by

p(1, 1, 1) = 1/8 , p(2, 1, 1) = 1/4 ,

p(1, 1, 2) = 1/8 , p(2, 1, 2) = 3/16 ,

p(1, 2, 1) = 1/16 , p(2, 2, 1) = 0,

p(1, 2, 2) = 0, p(2, 2, 2) = 1/4 , What is E[X|Y = 2]? What is E[X|Y = 2, Z = 1]?

In probability and statistics, the conditional expectation is the expected value of a random variable given specific conditions. In this blog post, we will solve the conditional expectations of X, Y, and Z, given their joint probability mass function.

Calculating E[X|Y = 2]:

To calculate the expected value of X given Y=2, we need to calculate the weighted average of the values of X, with their respective probabilities.

p(1, 2, 1) = 1/16, p(2, 2, 1) = 0

p(1, 2, 2) = 0, p(2, 2, 2) = 1/4

The expected value of X given Y=2 is 1/42 + 1/41 = 3/4.

Calculating E[X|Y = 2, Z = 1]:

To calculate the expected value of X given Y=2 and Z=1, we need to calculate the weighted average of the values of X, given their respective probabilities.

p(1, 2, 1) = 1/16, p(2, 2, 1) = 0

The expected value of X given Y=2 and Z=1 is 1/16*1 + 0 = 1/16.

We have solved the conditional expectations of X, Y, and Z, given their joint probability mass function. We have calculated the expected value of X given Y=2 and the expected value of X given Y=2 and Z=1. Understanding conditional expectations is important in solving real-world problems, as it helps us make predictions and decisions based on the given conditions.