Question 1172266: If the moment generating function for the random variable X is MX(t)=1/(1+t), what is the third moment of X about the point x=2?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down how to solve this problem.
**1. Identify the Distribution**
* The moment generating function (MGF) MX(t) = 1/(1 + t) is characteristic of an exponential distribution with parameter λ = 1, but with a slight twist.
* Let's rewrite it as MX(t) = 1/(1 - (-t)). This indicates that X is an exponential distribution with parameter λ = 1, but the MGF is defined for negative t.
* However, if we are to use this MGF to generate moments, we will treat it as 1/(1- (-t)) and proceed.
**2. Find the Moments about Zero**
* We can find the moments about zero by taking derivatives of the MGF and evaluating them at t = 0.
* First, rewrite the MGF as MX(t) = (1 + t)^(-1).
* **First moment (E[X]):**
* MX'(t) = -1(1 + t)^(-2)
* E[X] = MX'(0) = -1(1 + 0)^(-2) = -1
* **Second moment (E[X^2]):**
* MX''(t) = 2(1 + t)^(-3)
* E[X^2] = MX''(0) = 2(1 + 0)^(-3) = 2
* **Third moment (E[X^3]):**
* MX'''(t) = -6(1 + t)^(-4)
* E[X^3] = MX'''(0) = -6(1 + 0)^(-4) = -6
**3. Find the Third Moment about x = 2**
* We want to find E[(X - 2)^3]. We can expand this:
* E[(X - 2)^3] = E[X^3 - 6X^2 + 12X - 8]
* E[(X - 2)^3] = E[X^3] - 6E[X^2] + 12E[X] - 8
* Substitute the moments about zero we found:
* E[(X - 2)^3] = -6 - 6(2) + 12(-1) - 8
* E[(X - 2)^3] = -6 - 12 - 12 - 8
* E[(X - 2)^3] = -38
**Answer**
The third moment of X about the point x = 2 is -38.
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