Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $X_{1},\dots,X_{m}$ is iid $N(\mu_{1},\sigma^{2})$ and $Y_{1},\dots,Y_{n}$ is iid $N(\mu_{2},\sigma^{2})$. Is it true that the pooled estimate of the variance, $S_{p}^{2}$, has the property $\frac{(n-1)S_{p}^{2}}{\sigma^{2}}\sim\chi^{2}_{n-1}$, as is the case for a single sample variance $S^{2}$? I know that $S_{p}^{2}$ multiplied by a constant has a $\chi^{2}$ distribution, but am not sure if this property holds here.

share|improve this question
    
No: It's $\chi^2_{m+n-2}$. I'll post an answer below. –  Michael Hardy Jul 16 '13 at 22:05

1 Answer 1

up vote 1 down vote accepted

You omitted to say that the sequence $X_1,\ldots,X_m$ is independent of $Y_1,\ldots, Y_n$. I will take that to be assumed below.

I will take $S_p^2$ to be defined as follows: $$ S_p^2 = \frac{\sum_{i=1}^m (X_i-\bar X)^2 + \sum_{i=1}^n (Y_i-\bar Y)^2}{n+m-2} $$ where $\bar X$ and $\bar Y$ are the sample means.

Then $$ \frac{\sum_{i=1}^m (X_i-\bar X)^2}{\sigma^2} \sim \chi^2_{n-1} $$ and $$ \frac{\sum_{i=1}^n (Y_i-\bar Y)^2}{\sigma^2} \sim \chi^2_{m-1}. $$

The sum of two chi-square random variables that are independent of each other is a chi-square random variable, and the number of its degrees of freedom is just the sum of the numbers of degrees of freedom of the two terms in the sum.

Apply that to $(n+m-2)S_p^2$ (not to $(n-1)S_p^2$ as your question states).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.