SOLUTION: The first three terms of the power series expansion of the moment generating function of the random variable X are 1−t+{{{ t^2 }}}. What are the first three terms in the power

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Question 1197966: The first three terms of the power series expansion of the moment generating
function of the random variable X are 1−t++t%5E2+. What are the first three terms in the power
series expansion of the moment generating function of the random variable Y = 1 − X?
Answer by onyulee(41) About Me  (Show Source):
You can put this solution on YOUR website!
**1. Find the MGF of Y**
* **Definition of MGF for Y:**
* M_Y(t) = E[e^(tY)]
* Since Y = 1 - X, we have:
* M_Y(t) = E[e^(t(1-X))]
* M_Y(t) = E[e^t * e^(-tX)]
* M_Y(t) = e^t * E[e^(-tX)]
* **Recognize E[e^(-tX)]**
* E[e^(-tX)] is the MGF of -X, which is M_X(-t)
* **Therefore:**
* M_Y(t) = e^t * M_X(-t)
**2. Find the first three terms of M_X(-t)**
* Given M_X(t) = 1 - t + t^2, we can find M_X(-t) by substituting -t for t:
* M_X(-t) = 1 - (-t) + (-t)^2
* M_X(-t) = 1 + t + t^2
**3. Find the first three terms of M_Y(t)**
* M_Y(t) = e^t * M_X(-t)
* M_Y(t) = e^t * (1 + t + t^2)
* **Recall the Taylor series expansion of e^t:**
* e^t = 1 + t + (t^2)/2! + (t^3)/3! + ...
* **Multiply e^t with (1 + t + t^2) and consider only the first three terms:**
* M_Y(t) ≈ (1 + t + (t^2)/2) * (1 + t + t^2)
* M_Y(t) ≈ 1 + t + t^2 + t + t^2 + (t^2)/2
* M_Y(t) ≈ 1 + 2t + (5/2)t^2
**Therefore, the first three terms in the power series expansion of the moment generating function of the random variable Y = 1 - X are 1 + 2t + (5/2)t^2.**