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Compute the density function of Z = X2/X1 given that X2 and X1 both have the normal distribution with mean 0, variance 1 as their density.

I got this far:

$$ F(z) = P( X_2 \leq X_1 z) = \int\int \frac{1}{2 \pi} e^{-x^2-y^2} dx dy $$ change of variables $$ \int^{+\infty}_0 \int^{\tan^{-1}(z + 2\pi)}_{\tan^{-1}( z + \pi)} \frac{1}{2\pi}e^{-r^2}r dr d\theta = 1/2 $$

Now the textbook answer is $tan^{-1} z / \pi$ without any explanations and I have trouble understanding it as I am trying to self learn probability. Any help greatly appreciated.

Chen Ee Woon
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  • On the one hand, whomever voted down the question should provide an explanation as to why. On the other, your second equality, how did you get it? – William M. Nov 21 '18 at 18:51
  • I thought the goal is to find an expression for $F(Z \leq z)$ which is equivalent to $F(\frac{X_2}{X_1} \leq z)$ equivalent to $F(X_2 \leq z X_1)$. z is a real number. edit because I screwed up the math typesetting – Chen Ee Woon Nov 21 '18 at 18:54
  • You are probably given that $X_1$ is independent of $X_2$. Then one of the many posts here that has an answer to this question: https://math.stackexchange.com/questions/77873/how-calculate-the-probability-density-function-of-z-x-1-x-2?noredirect=1&lq=1. – StubbornAtom Nov 21 '18 at 18:58
  • Many thanks. I made a blunder multiplying by a negative number without changing the inequality. And I didn't realise Will is hinting at it. – Chen Ee Woon Nov 21 '18 at 19:15

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