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enter image description here I know this will be a f distribution.But it's not f(m,n) since the square sign is outside the summation.So it will be f(1,n).But i can't seem to know how exactly.

kjetil b halvorsen
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Abhisekkkk
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1 Answers1

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To answer your question, you must remember that:

  • if $X\sim N(\mu_x, \sigma^2_x)$ and $Y \sim N(\mu_y, \sigma^2_y)$ (independent) then $X + Y \sim N(\mu_x+\mu_y, \sigma^2_x + \sigma^2_y)$
  • if $X \sim N(\mu, \sigma)$ then $\frac{X-\mu}{\sigma} \sim N(0, 1)$
  • if $X\sim N(0, 1)$ then $\sum_{i=1}^n X_i^2 \sim \chi^2_n$
  • if $X \sim \chi^2_a$ and $Y \sim \chi^2_b$ are independent, then $\frac{\frac{X}{a}}{\frac{Y}{b}} \sim F_{a,b}$
ocram
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  • Perhaps another property to remember in order to reach the answer is that if $Z\sim \mathcal{N}(0,1)$ then $X = \mu + \sigma Z \sim \mathcal{N}(\mu,\sigma^2)$ – Emil Feb 08 '18 at 09:37
  • @Emil: Yes, indeed ! I had to do the exercice to figure this out ! Thank you ! I have added it (2nd bullet) – ocram Feb 08 '18 at 10:39
  • Same here! I had to solve it as well, and at first I misread the denominator having the same form as the nominator, i.e $\left(\sum_{j=1}^{n} Y_j\right)^2$, which led me to the conclusion of $W\sim F(1,1)$ as a ratio of two $\Gamma(\frac{1}{2}, mn)$ variables. Obviously, this was not the result of interest here, but I actually found this conclusion rather interesting, i.e. that $\forall m,n\in \mathbb{N}_{>0}$, $$ W := \dfrac{n\left(\sum_{i=1}^m X_i\right)^2}{m\left(\sum_{j=1}^n Y_j\right)^2} \sim F(1,1)$$ – Emil Feb 08 '18 at 17:22
  • also, I just noticed on your 2nd bullet point that you should change $\sigma$ to $\sigma^2$ in the distribution, otherwise people will be "standardizing" by the variance, not the std. deviation – Emil Feb 10 '18 at 20:09