# Thread: Problems of several random barialbles

1. ## Problems of several random barialbles

P(X<Y)= P(X<t and t<Y) ----> t is [0,1]
=P(X<t)P(t<Y)

P(X<t)=t;
since P(Y<t)= $t^2$
P(t<Y)=1- $t^2$

so P(X<Y)= t(1- $t^2$)
am i right???? or the answer should be a numerical value?

Question2

(a)
i know that i have to find P(Z<t), then differentiate P(Z<t) to obtain
the density function of Z.

where P(Z<t)=P(min(X1,X2...Xn)<t)
=1-P(X1>t)P(X2>t)...P(Xn>t)
=1-[1-P(X<t)]^100-----@

finally use the density function of Z to find the 1st moment and 2nd moment ...
so as to find the mean and variance...

however , the process of differentiating @ is so complicated...
any one show the details...?

2. that didn't make sense, that t stuff doesn't make sense
yes, it's a constant, this is just calc 3

The joint density is $f(x,y)=2y$ on the unit square

so your probability is the double integral over the appropriate region...

$\int_0^1 \int_x^1 2ydy dx$ or the simpler $\int_0^1 \int_0^y 2ydx dy=2/3$ by inspection

3. Originally Posted by matheagle
that didn't make sense, that t stuff doesn't make sense
yes, it's a constant, this is just calc 3

The joint density is $f(x,y)=2y$ on the unit square

so your probability is the double integral over the appropriate region...

$\int_0^1 \int_x^1 2ydy dx$ or the simpler $\int_0^1 \int_0^y 2ydx dy=2/3$ by inspection
thanks, matheagle, i didnt realise they are indepedent..

what about question 2??

4. Originally Posted by kin
thanks, matheagle, i didnt realise they are indepedent..

what about question 2??
The differentiation requires nothing more than the chain rule, which you ought to have learned (an appropriate calculus background is assumed in this subject):

$g_1(t) = [1 - \Pr(X < t)]^{99} f(t)$

and you ought to know how to calculate Pr(X < t) using the pdf of X.