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I was asked to prove that $Z^2 \sim \chi _1 ^2$ where $Z = \min (X,Y)$. Below is my working out: $$ F_{\mathcal{Z}^2}(z) := P(\mathcal{Z}^2 < z)$$ $$ = P(\mathcal{Z}^2 < z, X<Y) + P(\mathcal{Z}^2 < z, Y<X) $$ $$= P(\mathcal{Z}^2 < z | X<Y)\cdot P(X<Y) + P(\mathcal{Z}^2 < z | Y<X)\cdot P(Y<X)$$ $$= P(X^2 < z)\cdot \frac 12 + P(Y^2 < z)\cdot \frac 12$$ $$ = P(X^2 < z)\cdot \frac 12 + P(X^2 < z)\cdot \frac 12$$ $$ = P(X^2 < z)$$ $$= F_{X^2}(z)$$

My concern with my working out is with the second step. I am not sure that it is correct to split up the probability into the 2 cases where $X<Y$ and $Y<X$.

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  • $\begingroup$ This was asked not long back. $\endgroup$ – StubbornAtom May 22 '18 at 7:20
  • $\begingroup$ @StubbornAtom could you please send a link to the question because I cant seem to find the question. Thanks $\endgroup$ – Johnson. W May 22 '18 at 8:07
  • $\begingroup$ math.stackexchange.com/q/2790910/321264. $\endgroup$ – StubbornAtom May 22 '18 at 9:35
  • $\begingroup$ @StubbornAtom thanks for the link. I see that a similar has been asked before, but I feel like my question is different since I am asking whether my specific solution is valid or not which is not covered in that question $\endgroup$ – Johnson. W May 22 '18 at 13:13

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