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Given $a_1, a_2,\ldots, a_5$ are i.i.d random variables with $N(0,4)$. Find the value of parameter $c$ such that $\frac{c(a_1+a_2+a_3)^2}{a_4^2+a_5^2}$ has $F$ distribution.

My attempt: Let $Z = a_1+a_2+a_3$. Since all $3$ variables are i.i.d rvs with $N(0,4)$, we have: $Z$ follows $N(0,12)$, or $\frac{Z}{2\sqrt{3}}$ follows $N(0,1)$. Thus, $\frac{Z^2}{12}$ follows $\ {\chi}_1^2$.

In addition, by a well-known theorem, $\frac{a_4^2+a_5^2}{4}$ follows $\ {\chi}_2^2$. Thus, $\frac{(a_1+a_2+a_3)^2/12}{(\frac{a_4^2+a_5^2}{4})/2}$ follows $F_{1,2}$, so $c = \fbox{$\frac{2}{3}$}$.

My question: Could someone please help verify if my solution above is correct? If not, please help point out where my mistake is, and I would really appreciate it.

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    Nobody minds helping me review my solution above?? I am honestly grateful to know whether I made a mistake in the above attempt or not. – 2017-02-08

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Your solution is completely correct.

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    Thank you so much for your help, NCh!! – 2017-02-09
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    could you please help me with another problem, as I got quite stuck on this problem for several days:p http://math.stackexchange.com/questions/2139156/total-hourly-profit-of-food-stand – 2017-02-13