Question

In parts (a)−(d) below, either use the information given to determine the distribution of the random variable, or show that the information given is not sufficient by describing at least two different random variables that satisfy the given condition. (a) X is a random variable such that MX(t) = e 6t 2 when |t| < 2. (b) Y is a random variable such that MY (t) = 2 2−t for t < 0.5. (c) Z is a random variable such that MZ(t) = [infinity] for t ≥ 5. (d) W is a random variable such that MW (2)

Answer 1

a.) X ≈ N (0, 12)

b.) Y ≈ exp(2)

c.) The given distribution is not defined.

d.) Not a standard form of any distribution .

Explanation:

Given - In parts (a)−(d) below, either use the information given to determine

the distribution of the random variable, or show that the information

given is not sufficient by describing at least two different random

variables that satisfy the given condition.

To find - (a) X is a random variable such that Mₓ(t) =
`e^{6t^(2) }` when |t| < 2.

(b) Y is a random variable such that
`M_(Y)`(t) =
`(2)/(2 - t)` for t < 0.5.

(c) Z is a random variable such that
`M_(Z)`(t) = ∞ for t ≥ 5.

(d) W is a random variable such that
`M_(W)`(2) = 2

Proof -

a.)

As we know that the moment generating function of normal distribution is given by

Mₓ(t) =
`e^{ut + (1)/(2)t^(2) \sigma^(2)}`

Now,

Mₓ(t) =
`e^{6t^(2) }`

=
`e^{0(t) + (12)/(2) t^(2) }`

⇒μ = 0, σ = 12

Nd it is represented by

X ≈ N (0, 12)

b.)

Moment generating function of exponential function is given by

`M_(Y)`(t) =
`( 1 - (t)/(\theta))^(-1)`

Now,

`M_(Y)`(t) =
`(2)/(2 - t)` for t < 0.5.

=
`(1)/((2 - t)/(2) )`

=
`(1)/(1 - (t)/(2) )`

=
`( 1 - (t)/(2) )^(-1)`

And it is represented by

Y ≈ exp(2)

c.)

Z is a random variable such that
`M_(Z)`(t) = ∞ for t ≥ 5.

The given distribution is undefined.

d.)

`M_(W)`(2) = E(
`e^(2w)` )

=
`\int\limits^(-\infty)_(\infty) {e^(2w) f(w) } \, dw`

= 2

It is not the standard form of any distribution.