2
$\begingroup$

Suppose $X_1, X_2,$ and $X_3$ are independent and uniformly distributed in $[0,1]$.

(a) Find the joint PDF of $Y=\min\{{X_1,X_2,X_3}\}$ and $Z=\max\{{X_1,X_2,X_3}\}$.
(b) Determine $E(Z|Y)$.
(c) Verify the law of total variance in this setting by explicitly computing all terms.

(a) Using the answer below the joint PDF is $$f_X,_Y(x,y)=6(z-y)$$ I understand how the marginal pdfs are obtained below, but I am still struggling with finding the bounds of integration for the joint pdf.

(b) Using the solution from part (a) I was able to find $E(Z|Y)$. $$E(Z|Y)=\int_{-\infty}^\infty z f_{Z|Y} (z,y)dz$$
The conditional can be expressed as the joint divided by the marginal of Y, both of which have been determined and so the integral becomes

$$=\frac{1}{3(1-y^2)}\int_y^1z(6(z-y))dz$$ $$E(Z|Y)=\frac{2}{3}+\frac{y}{3}$$

(c) I must verify the formula $$var(z)=E(var(Z|Y))+var(E(Z|Y))$$ using part (b) $var(E(Z|Y))=var(\frac{2}{3}+\frac{y}{3})=\frac{1}{9}var(Y )$

here I am uncertain about $var(Y)$ I know that $Y=\min\{{X_1,X_2,X_3}\}$ but how do I find the var of this? For a single uniform random variable I know that the variance will be $\frac{1}{12}$ but since there are 3 $(X_1,X_2,X_3)$ should I multiple $\frac{1}{12}$ by 3?

now for $E(var(Z|Y))$, I am not sure about how to find the $var(Z|Y)$

If someone could please help clarify my questions, it would be very helpful to me.

$\endgroup$
9
  • $\begingroup$ Is there more information given about $X_1, X_2, X_3$? $\endgroup$
    – hessian
    Apr 22, 2017 at 23:56
  • $\begingroup$ @ElizabethHan Yes, I just updated the question $\endgroup$
    – user435438
    Apr 22, 2017 at 23:58
  • $\begingroup$ Also, X and Y are not independent, and thus you might not be able to separate their probability density functions. And actually, what is X? We only have $X_1$, $X_2$, etc. $\endgroup$
    – Jay Zha
    Apr 22, 2017 at 23:59
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/54505/… $\endgroup$ Apr 23, 2017 at 0:42
  • 1
    $\begingroup$ The direct approach to this is to note that $$[y\leqslant Y,Z\leqslant z]=\bigcap_{i=1}^3[y\leqslant X_i\leqslant z]$$ hence, for every $y<z$, $$P(y\leqslant Y,Z\leqslant z)=P(y\leqslant X\leqslant z)^3=(F_X(z)-F_X(y))^3$$ from which, differentiating twice, the joint PDF follows as $$f_{Y,Z}(y,z)=-\frac{\partial^2}{\partial y\partial z}P(y\leqslant Y,Z\leqslant z)=6f_X(y)f_X(z)(F_X(z)-F_X(y))\mathbf 1_{y<z}$$ $\endgroup$
    – Did
    Apr 23, 2017 at 9:44

2 Answers 2

1
$\begingroup$

The following may get you started, by finding the density functions of $Y$ and $Z.$

For $Z = \max(X_1, X_2, X_3),$ we have the CDF $$F_Z(t) = P(Z \le t) = P(X_1 \le t)P(X_2 \le t)P(X_3 \le t) = t^3,$$ for $0 <t <1.$ [If the maximum is less than $t,$ then so must each of the three independent $X_i$'s be less than $t.$] So $f_Z(t) = 3t^2,$ which is the density function of the distribution $\mathsf{Beta}(3,1).$

Similarly, for the minimum, $F_Y(t) = (1-t)^3,$ for $0 < t < 1,$ and $Y \sim \mathsf{Beta}(1,3).$

For illustration, the plot below shows histograms of simulations of 100,000 realizations of $Y$ and of $Z,$ along with the appropriate Beta densities. enter image description here

$\endgroup$
7
  • $\begingroup$ Hmmm, finding the joint PDF definitely requires another idea -- which you do not touch upon. $\endgroup$
    – Did
    Apr 23, 2017 at 9:40
  • 1
    $\begingroup$ @Did. Hmmm. In view of the suggested link, I didn't intend to give the complete answer. If you think that's appropriate, go ahead. $\endgroup$
    – BruceET
    Apr 23, 2017 at 13:55
  • $\begingroup$ Again: of course, incomplete answers (aka hints) are allright provided they are bearing on the question asked, the trouble here is that you solve an entirely different question and that the tools used to solve the latter are not the ones necessary to solve the former. $\endgroup$
    – Did
    Apr 24, 2017 at 6:21
  • $\begingroup$ This is a fairly standard problem. Perhaps this is a more directly relevant link on our site. There it is shown that $f(z,y) = 6(z-y),$ for $0 < y < z < 1.$ Integrating to get the marginals, one finds the beta distributions shown in my Answer. $\endgroup$
    – BruceET
    Apr 24, 2017 at 16:45
  • $\begingroup$ Yes, this is fairly standard, as at least 95% of the questions asked on the site. Re the method to solve the question, yes it is standard as well (and you might wish to read my comment from yesterday on main about this) but the trouble is that you do not explain it. $\endgroup$
    – Did
    Apr 24, 2017 at 17:38
0
$\begingroup$

I'll provided a way, might not be the most straight-forward/easy one, but should be able to solve it. Plus it starts from direct definitions, and could apply generically to statistics other than max/min as well.

$$F_{Y,Z}(y,z)=\mathbb P((Y\le y)\cap(Z\le z))$$ $$=\mathbb P(((X_1 \le y)\cup (X_2 \le y) \cup (X_3 \le y)) \cap ( (X_1 \le z) \cap (X_2 \le z) \cap (X_3 \le z)))$$ Now for the set: $$((X_1 \le y)\cup (X_2 \le y) \cup (X_3 \le y)) \cap ( (X_1 \le z) \cap (X_2 \le z) \cap (X_3 \le z))$$ $$=\bigcup_{i \ne j \ne k \in \{1,2,3\}}((X_i \le \min(y,z))\cap(X_j \le z) \cap (X_k \le z))$$

Then use the fact that $\mathbb P(A \cup B \cup C)= \mathbb P(A) + \mathbb P(B) + \mathbb P(C) - \mathbb P (A \cap B) - \mathbb P (C \cap B) - \mathbb P (A \cap C) + \mathbb P(A\cap B \cap C)$

to cut them into probability of intersections of events that's from the different $X_1,X_2,X_3$, and since the three r.v. are independent, the events from $\sigma(X_1), \sigma(X_2), \sigma(X_3)$ respectively are independent, then you could break up those intersection events into multiplication, and find the distribution.

$\endgroup$
3
  • $\begingroup$ To whoever down-voted, I think I am at least correct. Please tell me if you find this is wrong. Plus I said it is not the easiest one, but I start from the definition, and deduce by set operations $\endgroup$
    – Jay Zha
    Apr 23, 2017 at 1:00
  • $\begingroup$ And I think the value of this solution is that you do not think about any high-level trick and business logics, but you deduce from the very fundamental ground and rudimentary definitions about probability. I do not see a reason for a down vote just because this is not a smart solution. $\endgroup$
    – Jay Zha
    Apr 23, 2017 at 1:08
  • $\begingroup$ I did not down-vote, but I guess someone thinks this is unnecessarily complicated. Maybe OP will find it helpful, and up-vote. $\endgroup$
    – BruceET
    Apr 23, 2017 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy