# Conditional Prob -cont random variable

hello all

I have been workin on some problems involving conditional probability and continuous random variables and the thing is i dont know if i get the limits correct, anyway here is the problem, check it out, any suggestions would be helpful

$$f(y_1,y_2) =\left\{\begin{array}{cc}2,&\mbox{ if } 0\le y_1\le 1, 0\le y_2\le 1, y_1+y_2\le 1\\0, & \mbox{elsewhere}\end{array}\right$$

what i want to find was
$$P(Y_1\ge \frac{1}{2}|Y_2\le \frac{1}{4})$$
$$=\frac{\int_{0}^{\frac{1}{4}} \int_{\frac{1}{2}}^{1-y_2} 2 dy_1 dy_2}{\int_{0}^{\frac{1}{4}} \int_{0}^{1-y_2} 2 dy_1 dy_2}=\frac{3}{7}$$

also I wanted to find

$$P(Y_1\ge \frac{1}{2}|Y_2=\frac{1}{4})$$
$$=\frac{\int_{\frac{1}{2}}^{\frac{3}{4}} 2 dy_1}{\int_{0}^{1} 2 dy_1}=\frac{1}{4}$$
about the 3/4 that is where the intersection occurs

now have i got my limits correct? how do i know if i have the limits correct? are my answers corrrect?

steven

If you draw the triangle then the limits are simple. Your first problem becomes a ratio of areas and the second problem becomes a ratio of line segments. You will need to change the upper limit of the denominator of the second problem.

hello there

thanxs for that its an interesting way of looking at it, but I didnt get which areas and which segments should be put into ratio, so i have provided a diagram of my triangle, please point out which areas and segments are you refering to. from my understanding

$$P(Y_1\ge \frac{1}{2}|Y_2\le \frac{1}{4})$$

$$=\frac{D}{A+B+C+D}=\frac{3}{16}$$

$$P(Y_1\ge \frac{1}{2}|Y_2=\frac{1}{4})$$

$$=\frac{ys}{xs}=1/3$$

is this what you mean?

so was my answer for the first problem correct?
and how do i determine my limits for the second problem?

steven

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Yes, except P(Y1>=0.5| Y2<=0.25)= D/(D+B)=3/7. The upper limit of integration in the denominator of your 2nd problem should be 3/4. xs=3/4 and ys=1/4 and
P(Y1>=0.5|Y2=0.25)=(2*ys)/(2*xs)=1/3. Nice diagram.

Actually P(Y1>=0.5| Y2<=0.25)= (2*D)/(2*(D+B))=3/7.