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?
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.
To determine whether Y and Z are identically distributed, we need to compare their probability distributions.
Let's first find the probability distribution of Y, which represents the sum of the two dice tosses.
To calculate the probability distribution of Y, we need to consider all possible values that Y can take on and their respective probabilities.
The possible values for Y range from 2 (when both dice show 1) to 12 (when both dice show 6). We can compute the probability of each possible sum Y by considering all valid combinations of the dice outcomes.
Let's create a table to represent the probability distribution of Y:
Y | Probability
-------------------
2 | 1/36
3 | 2/36
4 | 3/36
5 | 4/36
6 | 5/36
7 | 6/36
8 | 5/36
9 | 4/36
10 | 3/36
11 | 2/36
12 | 1/36
Now let's consider the random variable Z, which is defined as Z = 2 * X.
Since X represents the outcome of a single dice toss, its values range from 1 to 6.
We can create a table to represent the probability distribution of Z:
Z | Probability
------------------
2 | 1/6
4 | 1/6
6 | 1/6
8 | 1/6
10 | 1/6
12 | 1/6
Comparing the probability distributions of Y and Z, we can see that they are not identical. The distribution of Y represents the sum of two dice tosses, whereas the distribution of Z represents the result of doubling a single dice toss. Therefore, the probabilities associated with the different values of Y and Z are not the same, and hence Y and Z are not identically distributed.
To determine if Y and Z are identically distributed, we need to compare the probability distributions of Y and Z.
For Y, we need to find the probability distribution of the sum of two dice tosses. This can be done by considering all possible outcomes:
Outcome of first dice toss (X1) | Outcome of second dice toss (X2) | Sum (Y)
1 1 2
1 2 3
1 3 4
1 4 5
1 5 6
1 6 7
2 1 3
2 2 4
2 3 5
2 4 6
2 5 7
2 6 8
3 1 4
3 2 5
3 3 6
3 4 7
3 5 8
3 6 9
4 1 5
4 2 6
4 3 7
4 4 8
4 5 9
4 6 10
5 1 6
5 2 7
5 3 8
5 4 9
5 5 10
5 6 11
6 1 7
6 2 8
6 3 9
6 4 10
6 5 11
6 6 12
From the above outcomes, we can calculate the probabilities for each sum Y. The probability distribution for Y is:
Sum (Y) | Probability
2 | 1/36
3 | 2/36
4 | 3/36
5 | 4/36
6 | 5/36
7 | 6/36
8 | 5/36
9 | 4/36
10 | 3/36
11 | 2/36
12 | 1/36
Now, let's consider the random variable Z = 2*X. Since X follows a uniform distribution with outcomes 1, 2, 3, 4, 5, 6 and probabilities 1/6 for each outcome, we can calculate the probability distribution for Z:
Value of X | Value of Z
1 | 2
2 | 4
3 | 6
4 | 8
5 | 10
6 | 12
From the above outcomes, we can calculate the probabilities for each value of Z. The probability distribution for Z is:
Value of Z | Probability
2 | 1/6
4 | 1/6
6 | 1/6
8 | 1/6
10 | 1/6
12 | 1/6
Comparing the probability distributions for Y and Z, we can see that they are not identically distributed. The probability distribution for Y is the distribution of sums of two dice tosses, while the probability distribution for Z is the distribution of two times the outcome of a single dice toss. Therefore, Y and Z have different probability distributions and are not identically distributed.