Question

let x be the value of the first die and y the sum of the values when two dice are rolled. compute the joint moment generating function of x and y

Answer 1

To find the z-score for y = 4 in a normal distribution with mean 2 and standard deviation 1, the formula (y - μ) / σ gives a z-score of 2.

Step-by-step explanation:

The question involves calculating the z-score for a given value of y when it is known that Y follows a normal distribution with a mean (μ) of 2 and a standard deviation (σ) of 1. The z-score is found using the formula:

z = (y - μ) / σ

For y = 4, the calculation would be:

z = (4 - 2) / 1

z = 2

This calculation tells us that a y value of 4 is 2 standard deviations above the mean of the distribution of Y.

Answer 2

The expression represents the joint moment generating function is:

`\[M_(X,Y)(t_1, t_2) = (1)/(6) \left(e^((t_1 + t_2)) + e^(2(t_1 + t_2)) + e^(3(t_1 + t_2)) + e^(4(t_1 + t_2)) + e^(5(t_1 + t_2)) + e^(6(t_1 + t_2))\right) * (1)/(6) \left(e^(t_2) + e^(2t_2) + e^(3t_2) + e^(4t_2) + e^(5t_2) + e^(6t_2)\right)\]`

The joint moment generating function (MGF) for two random variables
`\(X\)` and
`\(Y\)` is given by:

`\[M_(X,Y)(t_1, t_2) = E[e^(t_1 X + t_2 Y)]\]`

Here,
`\(X\)` is the value of the first die
`(\(X = x\))` and
`\(Y\)` is the sum of the values when two dice are rolled
`(\(Y = x + z\))`, where
`\(z\)` is the value of the second die).

Let's find the joint MGF step by step:

The MGF of a single die roll, denoted as
`\(M_X(t)\)`, is given by:

`\[M_X(t) = E[e^(tX)]\]`

For a fair six-sided die, the probability mass function (PMF) is
`\(P(X = k) = (1)/(6)\)` for
`\(k = 1, 2, 3, 4, 5, 6\)`. Therefore:

`\[M_X(t) = \sum_(k=1)^6 e^(tk) \cdot (1)/(6)\]`

`\[M_X(t) = (1)/(6) \left(e^t + e^(2t) + e^(3t) + e^(4t) + e^(5t) + e^(6t)\right)\]`

Now, let's express
`\(Y\)` in terms of
`\(X\)` and
`\(Z\)` (the second die):

`\[Y = X + Z\]`

So, the joint MGF for
`\(X\)` and
`\(Y\)` becomes:

`\[M_(X,Y)(t_1, t_2) = E[e^(t_1 X + t_2 (X + Z))]\]`

`\[M_(X,Y)(t_1, t_2) = E[e^((t_1 + t_2)X + t_2 Z)]\]`

As the dice are independent, the random variables
`\(X\)` and
`\(Z\)` are independent. Therefore, the joint MGF can be expressed as the product of individual MGFs:

`\[M_(X,Y)(t_1, t_2) = E[e^((t_1 + t_2)X)]\cdot E[e^(t_2 Z)]\]`

`\[M_(X,Y)(t_1, t_2) = M_X(t_1 + t_2) \cdot M_X(t_2)\]`

Substituting
`\(M_X(t)\)` into the equation:

`\[M_(X,Y)(t_1, t_2) = (1)/(6) \left(e^((t_1 + t_2)) + e^(2(t_1 + t_2)) + e^(3(t_1 + t_2)) + e^(4(t_1 + t_2)) + e^(5(t_1 + t_2)) + e^(6(t_1 + t_2))\right) * (1)/(6) \left(e^(t_2) + e^(2t_2) + e^(3t_2) + e^(4t_2) + e^(5t_2) + e^(6t_2)\right)\]`

This expression represents the joint moment generating function
`\(M_(X,Y)(t_1, t_2)\)` for the random variables
`\(X\)` and
`\(Y\)`.