asked 132k views
0 votes
Suppose that a random variable Y has the following probability density function: 8 f() = y3e-2y for y > 0 Find the moment-generating function of Y. (Make sure to show sufficient work to derive the function.)

1 Answer

0 votes

Answer:

Explanation:

To find the moment-generating function (MGF) of a random variable Y, we need to follow these steps:

1. Start with the probability density function (PDF) of Y, which is given as:

f(y) = 8y^3e^(-2y) for y > 0

2. Recall that the MGF of Y is defined as the expected value of e^(tY), where t is a constant:

M(t) = E[e^(tY)]

3. To find the MGF, we need to calculate the expected value by integrating e^(ty) multiplied by the PDF:

M(t) = ∫(e^(ty) * f(y)) dy

4. Substitute the given PDF into the integral:

M(t) = ∫(e^(ty) * 8y^3e^(-2y)) dy

5. Simplify the integral:

M(t) = 8 ∫(y^3e^((ty-2y))) dy

= 8 ∫(y^3e^(-y(2-t))) dy

6. Now, let's focus on the exponent term inside the integral. We can rewrite it as:

-y(2-t) = -y*2 + yt

7. Rewrite the integral:

M(t) = 8 ∫(y^3e^(-y*2) * e^(yt)) dy

8. Notice that the integral ∫(y^3e^(-y*2)) represents the gamma function, which has the property:

∫(y^ne^(-ky)) dy = n! / k^(n+1)

9. Using this property, we can simplify the integral:

M(t) = 8 * 3! / (2^4) * ∫(y^3e^(-y*2)) dy

= 24 / 16 * ∫(y^3e^(-y*2)) dy

= 3 / 2 * ∫(y^3e^(-y*2)) dy

10. Finally, the MGF becomes:

M(t) = 3 / 2 * ∫(y^3e^(-y*2)) dy

And that's the final expression for the moment-generating function of Y

answered
User Gladys
by
7.8k points

No related questions found

Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.