Conditional Prob -cont random variable

AI Thread Summary
The discussion revolves around calculating conditional probabilities involving continuous random variables. The user presents two problems, seeking confirmation on the limits of integration and the correctness of their answers. A participant suggests visualizing the problem using a triangle to simplify the limits and ratios involved. Corrections are provided, indicating that the first probability should be calculated as 3/7 and the second's upper limit of integration should be adjusted to 3/4, leading to a final probability of 1/3. The conversation emphasizes the importance of accurately determining limits and understanding the geometric interpretation of the problems.
steven187
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hello all

I have been workin on some problems involving conditional probability and continuous random variables and the thing is i don't 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,&amp;\mbox{ if }<br /> 0\le y_1\le 1, 0\le y_2\le 1, y_1+y_2\le 1\\0, &amp; \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
 
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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 referring 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.
 
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