# 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)= $\displaystyle t^2$
P(t<Y)=1-$\displaystyle t^2$

so P(X<Y)= t(1-$\displaystyle t^2$)
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 $\displaystyle f(x,y)=2y$ on the unit square

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

$\displaystyle \int_0^1 \int_x^1 2ydy dx$ or the simpler $\displaystyle \int_0^1 \int_0^y 2ydx dy=2/3$ 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 $\displaystyle f(x,y)=2y$ on the unit square

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

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

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