Question 1178448
Absolutely, let's prove that U2 is a chi-square random variable with ν - ν1 degrees of freedom.

**Understanding the Problem**

* **U1:** Chi-square random variable with ν1 degrees of freedom (U1 ~ χ²(ν1)).
* **U = U1 + U2:** Chi-square random variable with ν degrees of freedom (U ~ χ²(ν)).
* **U1 and U2:** Independent random variables.
* **Goal:** Prove that U2 ~ χ²(ν - ν1).

**Proof**

1.  **Moment Generating Functions (MGFs)**
    * We'll use MGFs because they uniquely characterize distributions.
    * The MGF of a chi-square random variable with k degrees of freedom is:
        * M_χ²(t) = (1 - 2t)^(-k/2), for t < 1/2.

2.  **MGFs of U1 and U**
    * Since U1 ~ χ²(ν1), its MGF is:
        * M_U1(t) = (1 - 2t)^(-ν1/2)
    * Since U ~ χ²(ν), its MGF is:
        * M_U(t) = (1 - 2t)^(-ν/2)

3.  **MGF of U2**
    * We know U = U1 + U2.
    * Since U1 and U2 are independent, the MGF of their sum is the product of their MGFs:
        * M_U(t) = M_U1(t) * M_U2(t)
    * We want to find M_U2(t), so rearrange:
        * M_U2(t) = M_U(t) / M_U1(t)
    * Substitute the known MGF's.
        * M_U2(t) = (1 - 2t)^(-ν/2) / (1 - 2t)^(-ν1/2)
    * Using the rules of exponents:
        * M_U2(t) = (1 - 2t)^(-ν/2 + ν1/2)
        * M_U2(t) = (1 - 2t)^(-(ν - ν1)/2)

4.  **Recognizing the Chi-Square MGF**
    * The resulting MGF, M_U2(t) = (1 - 2t)^(-(ν - ν1)/2), is exactly the MGF of a chi-square random variable with (ν - ν1) degrees of freedom.

5.  **Conclusion**
    * Since the MGF of U2 matches the MGF of a chi-square random variable with (ν - ν1) degrees of freedom, we conclude that:
        * U2 ~ χ²(ν - ν1)

**Therefore, U2 is a chi-square random variable with ν - ν1 degrees of freedom.**