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Let $X_1,\ldots, X_n$ be $n$ independent random variables uniformly distributed on $[0,1]$. Let be $Y=\min(X_i)$ and $Z=\max(X_i) $. Calculate the cdf of $(Y,Z)$ and verify $(Y,Z)$ has independent components.

Please help me out

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    $\begingroup$ Do you seriously think (Y,Z) is independent? $\endgroup$ – Did May 27 '14 at 21:11
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Even without computing anything, one can guess at the onset that $(Y,Z)$ is probably not independent since $Y\leqslant Z$ almost surely. Now, the most direct way to compute the distribution of $(Y,Z)$ might be to note that, for every $(y,z)$, $$ [y\lt Y,Z\leqslant z]=\bigcap_{i=1}^n[y\lt X_i\leqslant z], $$ hence, for every $0\lt y\lt z\lt1$, $$ P(y\lt Y,Z\leqslant z)=P(y\lt X_1\leqslant z)^n=(z-y)^n. $$ In particular, using this for $y=0$ yields, for every $z$ in $(0,1)$, $$ F_Z(z)=P(Z\leqslant z)=z^n. $$ Thus, for every $(y,z)$ in $(0,1)$, $$ F_{Y,Z}(y,z)=P(Y\leqslant y,Z\leqslant z) $$ is also $$ F_{Y,Z}(y,z)=P(Z\leqslant z)-P(y\lt Y,Z\leqslant z)=z^n-(z-y)^n\cdot\mathbf 1_{y\lt z}. $$ One may find more convenient to describe the distribution of $(Y,Z)$ by its PDF $f_{Y,Z}$, obtained as $$ f_{Y,Z}=\frac{\partial^2F_{Y,Z}}{\partial y\partial z}. $$ In the present case, for every $n\geqslant2$, $$ f_{Y,Z}(y,z)=n(n-1)(z-y)^{n-2}\mathbf 1_{0\lt y\lt z\lt1}. $$

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    $\begingroup$ Can't actually vote up, but thanks it was really usefull $\endgroup$ – White_noyze May 27 '14 at 22:09
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    $\begingroup$ @White_noyze I will vote up in the name of you. $\endgroup$ – Seyhmus Güngören May 27 '14 at 22:36
  • $\begingroup$ Out of curiosity, what does the notation $\mathbf{1}_{0<y<z<1}$ mean and how does this translate to being relevant here? No disrespect, just wondering, I am a student. @Did $\endgroup$ – afedder May 27 '14 at 22:56
  • $\begingroup$ Does this mean only for $y,z$ such that $0<y<z<1$ and $0$ otherwise? This would be my guess @Did $\endgroup$ – afedder May 27 '14 at 22:57
  • $\begingroup$ @afedder your guess is correct. $\endgroup$ – Seyhmus Güngören May 28 '14 at 19:27
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Obviously, $Y \leq Z$.

We use the facts that if $\min X_i \geq y$ then $X_1 \geq y, \ldots, X_n \geq y$ and similarly if $\max X_i \leq z$, $X_1 \leq z, \ldots, X_n \leq z$.


Now we calculate a joint distribution.

For $y \leq z$: $$P(Y \geq y, Z \leq z) = P(\{X_1, \ldots, X_n\} \geq y \text{ and } \{X_1, \ldots, X_n\} \leq z) = P( y \leq X_1, \ldots, X_n \leq z) = P(y \leq X_1 \leq z, y \leq X_2 \leq z, \ldots, y \leq X_n \leq z) = P(y \leq X_1 \leq z) P(y \leq X_2 \leq z) \ldots P(y \leq X_n \leq z)$$ where the last step follows from the independence of $X_1, \ldots, X_n$.

[If $y > z$, $P(Y \geq y, Z \leq z) = 0$ since $Y \leq Z$, so this completely specifies $P(Y \geq y, Z \leq z)$ and thus the distribution. ]

This specifies the joint distribution of $Y, Z$ (albeit a bit indirectly - we'd like $P(Y \leq y, Z \leq z)$, but we'll get to that next).


See the comments on this:

Now, show that you can factor $P(Y \geq y, Z \leq z)$ into a product of things that depends solely on $Y$ and solely on $Z$ to show independence (if they were independent) (so $P(Y \geq y, Z \leq z) = P(Y \geq y) P(Z \leq z)$ - you can find $P(Y \geq y)$ by setting $z=1$ in the expression for $P(Y \geq y, Z \leq z)$, and similarly find $P(Z \leq z)$ by setting $y=0$), and then from this independence, calculate the CDF as the product of the CDF's of $Y$ and $Z$ (since if events $A= \{ Y \geq y \}$ and $B = \{Z \leq z \}$ are independent, events $A$ and $B^C$ are also independent, so $P(Y \leq y, Z \leq z) = P(Y \leq y) P(Z \leq z)$).

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  • $\begingroup$ Do you seriously think (Y,Z) is independent? $\endgroup$ – Did May 27 '14 at 21:11
  • $\begingroup$ No, but if the person did the steps, they'd find that out when they got to the 3rd bar. $\endgroup$ – Batman May 27 '14 at 21:13
  • $\begingroup$ "and then from this independence, calculate the CDF as the product of the CDF's of Y and Z" ??? $\endgroup$ – Did May 27 '14 at 21:13
  • $\begingroup$ Happy? I'm trying to illustrate the general idea. $\endgroup$ – Batman May 27 '14 at 21:15
  • $\begingroup$ My happiness is irrelevant. The fact that the previous version of your post was misleading (and comforting the OP in their misconception) is relevant, yes. $\endgroup$ – Did May 27 '14 at 21:17
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Hint: What is the definition of the cumulative distribution function? The CDF of a random variable $X$ is defined as $$P(X \leq x) \,\,\,\text{for all $x \in \mathbb{R}$}\,\,.$$ Additionally, take advantage of the independence of the $X_i$ (as a set), however, be sure to note that $Y$ and $Z$ are not independent - this is clear, since the minimum of a set is dependent on the other elements of the set and similarly for the maximum of a set.

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  • $\begingroup$ I think Y,Z shouldn't be independent. I calculated $P(Z≤z)=z^n$ and $P(Y≤y)=1-(1-y)^n$. Then how do i get the cdf when $0≤y≤z≤1$? thanks $\endgroup$ – White_noyze May 27 '14 at 21:30
  • $\begingroup$ "I calculated $P(Z≤z)=z^n$ and $P(Y≤y)=1−(1−y)^n$". These are indeed the correct marginals, @user153633. $\endgroup$ – Did May 27 '14 at 21:48
  • $\begingroup$ Well, they should be correct. Anyway, what do you suggest? @afedder $\endgroup$ – White_noyze May 27 '14 at 21:49
  • $\begingroup$ can you help me with the joint distribution @Did ? $\endgroup$ – White_noyze May 27 '14 at 21:51
  • $\begingroup$ "Incorrect" was the wrong word, I apologize @White_noyze I only meant, I wouldn't use $P(Y \leq y)$, although you have computed it correctly...you should use $P(Z \leq z)$, however $\endgroup$ – afedder May 27 '14 at 22:43

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