Multi variable PDF Calculations

Sep 2012
838
89
Canada
Hey guys :) hope you're doing well !

I have this problem that has caught me off guard:

Let X be a continous random variable with PDF:

\(\displaystyle f_(xy) (x,y)= xc^2y ; 0 \leq y \leq x \leq 1 ; 0 else where\)

a) Find the range
I computed this to be the area bounded by the line y=x, x-axis and x=1.

b) Find the constant c
From the fact that the total area under the range must be 1 , I get c=10

c) Find the marginal PDFs, fx (x) and fy(y)
I get:

\(\displaystyle f_x(x) = \int_0^x 10x^2y dy = 5x^4 ; 0 \leq x \leq 1\)
\(\displaystyle f_y(y) = \int_0^y 10x^2y dx = 10/3 y^4 ; 0 \leq x \leq 1\)

d) Find \(\displaystyle P(Y\leq X/2)\)
I am having trouble here, I'm not sure I understand what the question is asking. What does 'Y' represent? Am I going to be working with the marginal PDF with respect to Y, or am I working with the multi variable original function?

Can I make the substitution y=x and solve using the marginal PDF for the variable y?

e) Find \(\displaystyle P (Y \leq X/4 | Y \leq X/2) \)
I am completely lost here. I would appreciate an explanation of what is going on.
 

HallsofIvy

MHF Helper
Apr 2005
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7,909
Initially, you are given that y must lie between 0 and x and that x must be no less than y and less than or equal to 1. In an xy-coordinate system that is the triangle with vertices (0, 0), (1, 0), and (1, 1). The exercise you ask about requires also that Y< X/2. y= x/2 is the line through (0, 0) and (1, 1/2). The original triangle has area 1/2 while the second triangle, with vertices (0, 0), (1, 0), and (1, 1/2), has area 1/4. The probability of a point being in the second triangle, given that it is in the first, is (1/4)/(1/2)= 1/2.
 
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Sep 2012
838
89
Canada
Yes, I agree with the domain of the PDF https://www.desmos.com/calculator/jjmrz24wf8

So, for part (d) , we are computing the probability with out using the PDF ? How can we get a probability from with out using the actual function, f(x,y)=10x^2y ?
If \(\displaystyle P(Y \leq X/2)\) is simply the restriction on the domain, then would it not make sense to evaluate:

\(\displaystyle \int_0^1 \int_0^{x/2} 10x^2y dy dx\)

Edit: So I just solved the integral above and I got P(Y<x/2) = 1/4 ! So, yes that is the case. But I still don't understand difference between big X and small x and part (e) !


Also, what is the difference between a big Y/X and a small y/x in this case?

For part (e), how do you mathematically define this, because I still don't get it.
 
Last edited:
Sep 2012
838
89
Canada
I attempted part (e). Here is my solution and logic.

Since the range of \(\displaystyle P(Y\leq X/2)\) contains in it the range defined by \(\displaystyle Y=X/4\), then for

\(\displaystyle Let A = Y \leq X/4\)
\(\displaystyle Let B = Y \leq X/2\)

\(\displaystyle P(A|B) =\frac{ P(A and B)}{P(B)}\)

Here the probability A and B is just \(\displaystyle P(A)\) because it is contained in P(B). Moreover, P(B) is as I calculated previously.

So, compute:

\(\displaystyle \int_0^1 \int_0^{x/4} 10x^2y dydx = 1/16\)

Finally,

\(\displaystyle P(A|B) =\frac{ P(A and B)}{P(B)} = (1/16) / (1/4) = 1/4\)

Is this correct?
I am not sure why in your solution, you are referencing the original range defined by y=x.

Thank you in advance !