Question 1164808
To find the sampling distribution and the moment-generating function (MGF) of the given statistic, we rely on the properties of independent normal random variables.

### 1. The Probability Distribution of the Sample

Since  is a random sample of size 4 from a **standard normal population**, each  is independent and identically distributed (i.i.d.) such that:



The joint probability distribution of the sample is the product of their individual normal density functions:


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### 2. The Sampling Distribution of the Statistic

Let the statistic be .
Note that for any , the square of the variable follows a **Chi-square distribution with 1 degree of freedom**:


However, our statistic  is a **weighted sum** of independent Chi-square variables:


Because the coefficients (weights) are not all equal to 1,  does **not** follow a standard Chi-square distribution. Instead, it follows a **Generalized Chi-square distribution** (specifically, a linear combination of Chi-squares). There is no simple named probability density function for this, but it is fully characterized by its MGF.

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### 3. Moment-Generating Function (MGF)

The MGF of a standard Chi-square variable  is:


Since the variables are independent, the MGF of a sum is the product of the individual MGFs. For a weighted variable , the MGF is .


Substitute the formula:


**Condition for Existence:** The MGF is defined only when all terms inside the square root are positive. The strictest constraint comes from , so .

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### Summary

* **Sample Distribution:** Jointly Normal .
* **Statistic Distribution:** A weighted sum of  variables (Generalized Chi-square).
* **MGF:**  for .

Would you like me to calculate the **mean** and **variance** of this statistic using its MGF?