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X~N(4, $\sigma^2$) and Y~N(1, $\sigma^2$) are independent.

$$A=\frac{\sum_{i=1}^N (X_i-E(X))^2}{\sum_{i=1}^M (Y_i-E(Y))^2}$$

Find the distribution of A?

I tried this way.

$$\frac{M-1}{N-1}A=\frac{\frac{1}{N-1}\sum_{i=1}^N (X_i-E(X))^2}{\frac{1}{M-1}\sum_{i=1}^M (Y_i-E(Y))^2}=\frac{S_x^2}{S_y^2}=\frac{(\frac{N-1}{\sigma^2})S_x^2/(N-1)}{(\frac{M-1}{\sigma^2})S_y^2/(M-1)}$$

So I concluded that $\frac{M-1}{N-1}A$ ~$F_N-_1,_M-_1$

But the question is about A, not about $\frac{M-1}{N-1}A$

I need some help! Please give me an advice.

Chloe
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1 Answers1

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I'm just going to assume you mean the $X_i \sim \mathcal{N}(4, \sigma^2)$ and $Y_i \sim \mathcal{N}(1, \sigma^2)$ are mutually independent.

Notice $X_i - E[X_i] \sim \mathcal{N}(0, \sigma^2)$ and $Y_i - E[Y_i] \sim \mathcal{N}(0, \sigma^2)$. Furthermore, $\dfrac{X_i - E[X_i]}{\sigma} \sim \mathcal{N}(0, 1)$ and similarly with the $Y_i$.

Notice that independence still holds with the $\dfrac{X_i - E[X_i]}{\sigma}$ and the $\dfrac{Y_i - E[Y_i]}{\sigma}$. It follows that $S_X = \sum_{i=1}^{N}\left(\dfrac{X_i - E[X_i]}{\sigma}\right)^2 \sim \chi^2_N$ (see here). Similarly, $S_Y=\sum_{i=1}^{M}\left(\dfrac{Y_i - E[Y_i]}{\sigma}\right)^2 \sim \chi^2_M$.

Notice that $\dfrac{S_X}{S_Y} = A$ (the constants $\sigma^2$ cancel out) and $S_X, S_Y$ are independent.

It is well-known that $\dfrac{S_X/N}{S_Y/M} = \dfrac{S_X}{S_Y}\dfrac{M}{N} \sim F_{N, M}$ (see here). So $\dfrac{S_X}{S_Y} \sim \dfrac{N}{M}F_{N, M} $, a scaled $F$ distribution.

Clarinetist
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