Question

Let X be a random variable that records the outcome of a dice toss. The probability distribution for X is as follows:
Outcome: 1, 2, 3, 4, 5, 6
Probability: 1/6, 1/6, 1/6, 1/6, 1/6, 1/6

Suppose now that we toss two dice. X1 records the outcome of the first toss, and X2 records the outcome of the second toss. We are interested in the sum of the two dice tosses,
Y= X1+ X2.

Let's define another new random variable Z such that
Z = 2∗X
Are Y and Z identically distributed?

Answer 1

No, Y and Z are not identically distributed.

Y represents the sum of the two dice tosses, ranging from 2 to 12. The probability distribution for Y can be calculated by considering all possible combinations of the dice outcomes and summing their probabilities. For example, to get a sum of 2, both dice must show a 1, which has a probability of (1/6) * (1/6) = 1/36. Similarly, to get a sum of 3, one dice must show a 1 and the other must show a 2, or vice versa, which has a probability of 2 * (1/6) * (1/6) = 1/18. Continuing this calculation for all possible sums, we can determine the probability distribution for Y.

Z, on the other hand, is defined as 2 times X, where X is the outcome of a single dice toss. Since X can take on values from 1 to 6, Z can take on values from 2 to 12. However, the probability distribution for Z is not the same as the probability distribution for Y. To calculate the probability distribution for Z, we need to consider each possible value of X and multiply it by 2. For example, the probability of getting Z = 2 is P(X = 1) * 2 = (1/6) * 2 = 1/3. Continuing this calculation for all possible values of X, we can determine the probability distribution for Z.

Since the probability distributions for Y and Z are different, Y and Z are not identically distributed.