SOLUTION: Let U1 and U2 be independent random variables. Suppose that U1 is χ^2 with ν1 degrees of freedom while U = U1 + U2 is chi-square with ν degrees of freedom, where ν

SOLUTION: Let U1 and U2 be independent random variables. Suppose that U1 is χ^2 with ν1 degrees of freedom while U = U1 + U2 is chi-square with ν degrees of freedom, where ν > ν1. Then

Question 1178448: Let U1 and U2 be independent random variables. Suppose that U1 is χ^2 with ν1 degrees of freedom while U = U1 + U2 is chi-square with ν degrees of freedom, where ν > ν1. Then prove that U2 is chi-square random variable with ν − ν1 degrees of freedom.
thank you
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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.**

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