Question 1197966
**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.**