Finding a probability given joint p.d.f of the continuous random variables

In summary, the conversation discusses a problem involving the joint p.d.f of two continuous random variables and finding the probability P(X+Y<2). The solution requires double integration, with the limits of integration determined by drawing a picture of the relevant area. This approach is similar to finding the area of a region in Calc III.
  • #1
nabilsaleh
14
0
I'm having a trouble doing this kind of problems :S

Lets try this for example:

The joint p.d.f of the continuous random variable X and Y is:

f(x,y)= (2y+x)/8 for 0<x<2 ; 1<y<2

now we're asked to find a probability, say P(X+Y<2)

I know i have to double integrate but how do I choose my integration limits and what do I do after I finish integrating?

My Attempt at the solution is:

P(X+Y<2) = \int_ \int (2y+x)/8 = 1 = \int [y^2/8 + xy/ 8]
 
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  • #2
Inner integral of x, (0,2-y), outer integral of y (1,2).

or

Inner integral of y, (1,2-x), outer integral of x (0,1). (For x > 1, y < 1).
 
  • #3
nabilsaleh said:
I'm having a trouble doing this kind of problems :S

Lets try this for example:

The joint p.d.f of the continuous random variable X and Y is:

f(x,y)= (2y+x)/8 for 0<x<2 ; 1<y<2

now we're asked to find a probability, say P(X+Y<2)

I know i have to double integrate but how do I choose my integration limits and what do I do after I finish integrating?

My Attempt at the solution is:

P(X+Y<2) = \int_ \int (2y+x)/8 = 1 = \int [y^2/8 + xy/ 8]

The trick to solving this kind of problem is to draw a picture of the relevant area. If you were in Calc III doing a double integral for area, that is what you would do to help you figure out the limits of integration. Same thing here. In Calc III if you were calculating the area of a region R you would do this integral:
[tex]\iint_R 1\,dxdy[/tex]
with appropriate limits for dxdy or dydx, whichever is easier. The only difference is that instead of integrating 1 and getting area, you integrate your joint density f(x,y) over the region. In your example the appropriate region is a triangle. More interesting would be if you wanted P(X+Y<3). Do you see that in that case you would to need to break the region into two parts as a dydx integral but not as a dxdy integral? Draw a picture and see.
 
Last edited:

Related to Finding a probability given joint p.d.f of the continuous random variables

What is a joint p.d.f of continuous random variables?

A joint p.d.f (probability density function) of continuous random variables is a mathematical function that describes the probability of a particular set of values occurring simultaneously for multiple continuous random variables. It is used to model the relationship between these variables and can be used to find the probability of a specific event or range of values occurring.

How is the joint p.d.f related to the individual p.d.f's of the continuous random variables?

The joint p.d.f is a combination of the individual p.d.f's of the continuous random variables. It takes into account the relationship between the variables and provides a more accurate representation of the probability of events occurring simultaneously.

How do you find the probability given a joint p.d.f of continuous random variables?

To find the probability given a joint p.d.f, you must integrate the joint p.d.f over the specified range of values. The result will be the probability of that event occurring for the given variables.

What is the difference between a joint p.d.f and a marginal p.d.f?

A joint p.d.f describes the relationship between multiple continuous random variables, while a marginal p.d.f describes the probability distribution of a single variable without considering the other variables. The marginal p.d.f can be obtained by integrating the joint p.d.f over all other variables.

Can a joint p.d.f be used for discrete random variables?

No, a joint p.d.f is only used for continuous random variables. For discrete random variables, a joint p.m.f (probability mass function) is used instead.

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