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Let $X \sim N (0, 1)$ and $Y ∼ N (0, 1)$ be two independent random variables, and define $Z = \min(X, Y )$. Prove that $Z^2\sim\chi^2(1),$ i.e. Chi-Squared with degree of freedom $1.$

I found the density functions of $X$ and $Y,$ as they are normally distributed. How would one use the fact that $Z = \min(X,Y)$ to answer the question? Thanks!

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    $\begingroup$ Maybe it would it be useful to express $Z$ as a liner combination of $X+Y$ and $|X-Y|$, using properties of maximum and minimum. $\endgroup$
    – AnyAD
    May 22, 2018 at 2:52
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    $\begingroup$ Can you clarify what the subscript $1$ in $X_1^2$ refers to? $\endgroup$ May 22, 2018 at 3:21
  • $\begingroup$ @Programmer : Your comment on $X_1$ as "the degree of freedom 1" does not make sense. That is because it is not clear what that means, and how $X_1$ would differ from $X$. I believe you are really trying to show that $Z^2$ has the same distribution as $X^2$. Thus, I believe the "1" subscript is just a typo. $\endgroup$
    – Michael
    May 22, 2018 at 3:46
  • $\begingroup$ I doubt the correctness of the conclusion... Because the p.d.f. of $Z^2$ can be explicitly figured out, and when I figured it out, it was not the one of $\chi^2_1$ variable... $\endgroup$
    – Wanshan
    May 22, 2018 at 4:11
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    $\begingroup$ @Wanshan If $X, Y$ are iid and continuous random variables with the property that $X$ and $-X$ have the same distribution, then $(\min[X,Y])^2$ has the same distribution as $X^2$. You can condition on events $\{X<0, Y<0\}, \{X<0, Y>0\}, \{X>0, Y<0\}, \{X>0, Y>0\}$. $\endgroup$
    – Michael
    May 22, 2018 at 4:13

3 Answers 3

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Let $\Phi$ be the standard normal cdf. First, find the cdf of $Z^2$. For any $z>0$,

$$ \begin{align} P(Z^2\le z) &=P(Z>-\sqrt{z})-P(Z>\sqrt{z})\\ &=P(X>-\sqrt{z})P(Y>-\sqrt{z})-P(X>\sqrt{z})P(Y>\sqrt{z})\\ &=(1-\Phi(-\sqrt{z}))^2-(1-\Phi(\sqrt{z}))^2. \\&=(\Phi(\sqrt{z}))^2-(1-\Phi(\sqrt{z}))^2. \\&=2\Phi(\sqrt z)-1 \end{align} $$

On the other hand, $$ P(X^2\le z)=P(X\le \sqrt{z})-P(X<-\sqrt{z})=\Phi(\sqrt{z})-\Phi(-\sqrt{z})=\Phi(\sqrt{z})-(1-\Phi(\sqrt{z})) $$ As you can see, $P(X^2\le z)=P(Z^2\le z)$ for all $z$, QED.

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  • $\begingroup$ I see this is same as existing answers, for some reason they had not loaded when I wrote this. $\endgroup$ May 22, 2018 at 14:26
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$1-F_Z(t) = P(Z>t) = P(X>t)P(Y>t) =\frac{1}{2\pi}\left[ \int_t^{\infty}\exp(-x^2/2) \, dx \right]^2$. Take derivative w.r.t. $t$ and we can get $$ f_Z(t) = -\frac{d}{dt}\frac{1}{2\pi} \left[ \int_t^\infty \exp(-x^2/2)\,dx \right]^2 = \frac{1}{\pi}\exp(-t^2/2)\left[\int_t^{\infty}\exp(-x^2/2)\,dx\right]. $$ Now let $W = Z^2$ \begin{align} 1-F_W(t) = {} & P(W>t) = P(Z>\sqrt{t})+P(Z<-\sqrt{t})\\[10pt] = {} & \int_{\sqrt{t}}^{\infty}\frac{1}{\pi}\exp(-s^2/2) \left[\int_s^\infty \exp(-x^2/2)\,dx\right]\,ds \\[10pt] & {} + \int_\infty^{-\sqrt{t}}\frac{1}{\pi}\exp(-s^2/2) \left[ \int_s^{\infty}\exp(-x^2/2)\,dx\right]\,ds\\[10pt] = {} & \int_{\sqrt{t}}^{\infty}\frac{1}{\pi}\exp(-s^2/2)\left[ \int_s^\infty \exp(-x^2/2)\,dx \right] \, ds \\[10pt] & {} + \int^\infty_{\sqrt{t}}\frac{1}{\pi}\exp(-s^2/2) \left[ \int^{s}_{-\infty}\exp(-x^2/2)\,dx\right]\,ds\\[10pt] = {} & \int_{\sqrt{t}}^\infty \frac{1}{\pi}\exp(-s^2/2)\frac{\sqrt{2\pi}}{2}\,ds \end{align} Taking derivative we can get $f_W(t) = \frac{1}{\sqrt{2\pi t}}\exp(-t/2)$, which is the same as $f_{\chi^2_1}(t) = \frac{1}{\sqrt{2\pi t}}\exp(-t/2)$.

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    $\begingroup$ @Michael Yes I just noticed that and fixed it. Thanks for your discussion! It's really helpful. $\endgroup$
    – Wanshan
    May 22, 2018 at 4:53
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    $\begingroup$ Since you've shown that $P(Z>z)=P(X>z)^2$, you could also just write $$P(Z^2>t)=P(Z>\sqrt{t})+P(Z<-\sqrt{t})\\ = P(Z>\sqrt{t})+1-P(Z>-\sqrt{t})\\ =P(X>\sqrt{t})^2 + 1-P(X>-\sqrt{t})^2\\ =P(X>\sqrt{t})^2+1-(1-P(X<-\sqrt{t}))^2\\ =2P(X>\sqrt{t}),$$ hence $P(Z^2\le t) = 1-2P(X>\sqrt{t})=P(-\sqrt{t}<X<\sqrt{t})=P(X^2<t).$ QED $\endgroup$
    – r.e.s.
    May 22, 2018 at 5:34
  • $\begingroup$ @r.e.s. Good idea! That's far more elegant. $\endgroup$
    – Wanshan
    May 22, 2018 at 13:31
  • $\begingroup$ @Programmer Yes I agree with you. Maybe he meant to write chi square distribution. But since X is a standard normal variable, it's also correct to say that $Z^2$ has the same distribution as $X^2$ :) $\endgroup$
    – Wanshan
    May 22, 2018 at 19:21
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Since $X$ and $Y$ are independent, we can see that $f(x,y) = f(x)f(y)$.

Now, if we transform our axis from $X,Y$ to $X_1,Y_1$, where $X_1 = \frac{1}{\sqrt{2}}(X+Y)$ and $X_2 = \frac{1}{\sqrt{2}}(X-Y)$ - essentially, we are rotating the $XY$ space by a +45 degrees.

Since the distribution $f(x,y)$ has radial symmetry, we can see that $f(x_1,x_2) =f(x,y) $. From this we can see that, even $X_2$ $∼ N (0, 1)$.

Now, the variable $Z = min(X,Y)$ lies to the space right of the line $\frac{1}{\sqrt{2}}(X-Y)$ in the $XY$ space.

Therefore, distribution of $Z^2$ is the same as the distribution of $X_2^{2}$ - which is essentially a Chi-square distribution with degrees of freedom = 1.

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