MHB Continuous joint probability density functions

AI Thread Summary
The discussion focuses on calculating the constant 'a' for a joint probability distribution function defined as a(x + y^2) within the bounds 0 <= x <= 2 and 0 <= y <= 2. The integral of this function must equal one to ensure it is a legitimate probability distribution. Participants discuss setting up the double integral and using tools like Integral Calculator to evaluate it, with one user suggesting the result is 28/3, leading to a proposed value of 3/28 for 'a'. Additionally, the conversation touches on deriving the marginal probability density function of Y and calculating the probability for the range 0.25 < Y < 0.75. The importance of understanding double integrals in this context is emphasized, indicating that familiarity with these concepts is crucial for solving such problems.
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Consider the following joint probability distribution function of (X , Y):

a(x + y^2) {0<=x<=2, 0<=y<=2}
0 otherwise

Calculate the value of the constant a that makes this a legitimate probability distribution. (Round your answer to four decimal places as appropriate.)

And then,
For the distribution above, determine the marginal probability density function of Y and use it to calculate the probability that 0.25 < Y < 0.75. (Round your answer to the fourth decimal place as appropriate.)
 
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What have you tried so far? What properties of joint probability distribution functions do you know? There is one property that was used in a thread you made yesterday which is useful here, just extended to two variables.
 
OK so this is what I've done:
I set up the integral from 0 to 2 of (x+y^2)dx dy = 1
I've never done an integral with 2 variables before but I plugged it into this calculator:
Integral Calculator - Symbolab
and it gave me 28/3 ? Set that = 1 and divide and get 3/28?

That's my best guess so far, but I could be way off.
 
You've never worked with double integrals before? Do you know how to solve something like this?

$$\int_{0}^{1} \int_{1}^{5} x+ydydx$$

If not you haven't been given the tools you need to solve this problem so it's totally understandable why this is difficult! :)

You are on the right track though in your reasoning. The double integral of the joint pdf should equal one:

$$\int_{0}^{2} \int_{0}^{2} a(x+y^2)dydx=1$$
 
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