Conditional Prob -cont random variable

In summary, the conversation is about problems involving conditional probability and continuous random variables. The speaker is unsure if they have the correct limits and asks for suggestions. Another person suggests looking at the triangle and changing the upper limit for the second problem. The speaker provides a diagram and the other person confirms their answers are correct.
  • #1
steven187
176
0
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

[tex]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[/tex]


what i want to find was
[tex]P(Y_1\ge \frac{1}{2}|Y_2\le \frac{1}{4})[/tex]
[tex]=\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}[/tex]

also I wanted to find

[tex]P(Y_1\ge \frac{1}{2}|Y_2=\frac{1}{4})[/tex]
[tex]=\frac{\int_{\frac{1}{2}}^{\frac{3}{4}} 2 dy_1}{\int_{0}^{1} 2 dy_1}=\frac{1}{4}[/tex]
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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  • #2
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.
 
  • #3
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

[tex]P(Y_1\ge \frac{1}{2}|Y_2\le \frac{1}{4})[/tex]

[tex]=\frac{D}{A+B+C+D}=\frac{3}{16}[/tex]

[tex]P(Y_1\ge \frac{1}{2}|Y_2=\frac{1}{4})[/tex]

[tex]=\frac{ys}{xs}=1/3[/tex]

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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  • #4
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.
 
  • #5
Actually P(Y1>=0.5| Y2<=0.25)= (2*D)/(2*(D+B))=3/7.
 

What is a conditional probability for a continuous random variable?

Conditional probability for a continuous random variable refers to the likelihood of an event occurring given that another event has already occurred. This probability is calculated using the formula P(A|B) = P(A and B) / P(B), where A and B are events and P(A and B) is the probability of both events occurring together.

How is conditional probability different for continuous random variables compared to discrete random variables?

Conditional probability for continuous random variables is calculated using integration, while for discrete random variables it is calculated using summation. This is because continuous random variables can take on an infinite number of values, while discrete random variables can only take on a finite number of values.

What is the relationship between conditional probability and independence?

If two events A and B are independent, then the conditional probability of event A given event B is equal to the unconditional probability of event A. In other words, the occurrence of event B does not affect the likelihood of event A occurring. However, if two events are not independent, then the conditional probability of event A given event B may be different from the unconditional probability of event A.

How can conditional probability be used in real-life situations?

Conditional probability is commonly used in fields such as statistics, economics, and machine learning to make predictions and decisions based on past events. For example, in medical diagnosis, conditional probability can be used to calculate the likelihood of a patient having a certain disease given their symptoms and medical history.

What are some common misconceptions about conditional probability for continuous random variables?

One common misconception is that the probability of a continuous random variable taking on a specific value is equal to 0. This is not true, as the probability of a continuous random variable taking on a specific value is actually determined by its probability density function. Another misconception is that conditional probability always follows the same rules as unconditional probability, which is not always the case for dependent events.

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