Transformation Of Probability Density Functionsby caffeine Tags: density, functions, probability, transformation 
#1
Jan2007, 10:46 PM

P: n/a

1. The problem statement, all variables and given/known data
Let X and Y be random variables. The pdfs are [itex]f_X(x)=2(1x)[/itex] and [itex]f_Y(y) = 2(1y)[/itex]. Both distributions are defined on [0,1]. Let Z = X + Y. Find the pdf for Z, [itex]f_Z(z)[/itex]. 2. Relevant equations I'm using ideas, not equations. 3. The attempt at a solution I'm dying of curiosity about where I'm going wrong. I'm so sure of each step, but my answer can't be correct because [itex]\int_0^2 f_Z(z)\,dz[/itex] is zero! Here's my logic. Consider the cdf (cumulative distribution function) for Z: [tex] F_Z(z) = P(Z\le z) = P(X+Y \le z) [/tex] Here, [itex]F_Z(z)[/itex] is the volume above the triangle shown in the image I attached to this message (in case something happens to the attachment, it's the triangle in quadrant 1 bounded by x=0, y=0 and x+y=z.) The volume above the shaded region represents [itex]F_Z(z)[/itex]. [tex] F_Z(z) = \int_{x=0}^{x=z} 2(1x)\int_{y=0}^{y=zx} 2(1y)\,dy\,dx [/tex] Performing the integrals gives [itex]F_Z(z) = \frac{1}{6}z^4  \frac{4}{3}z^3 + 2z^2[/itex]. Then taking the derivative of the cdf gives the pdf: [tex] f_Z(z) = \partial_z F_Z(z) = \frac{2}{3}z^3  4z^2 +4z [/tex] Unfortunately, this can't be right because the integral of this function over [0,2] gives zero. I also would've expected that the maximum of [itex]f_Z(z)[/itex] would be at z=0 since individually, X and Y are most likely to be zero. I checked my algebra and calculus with Mathematica; it looked fine. I think there's a conceptual problem. There must be something I don't understand or some point I'm not clear about. What did I do wrong? 



#2
Jan2107, 01:03 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

You have to split the problem into different cases.
i) z<0 ii) z<1 iii) 1<z<2 iv) z>2 In each case the area of integration is different. But for each case, your model for the integral is [tex]F_Z(z)=\int\int_{\{(x,y)\in [0,1]\times[0,1]:\ y\leq zx, \}}f_{XY}(x,y)dxdy[/tex] So you're integrating over the area that's the intersection of the square [0,1] x [0,1] with the area under the curve y=zx. (Btw, you never said that X and Y are independant but I assume they are?) 


#3
Jan2107, 10:57 AM

P: n/a

Ahhh.... yes. I *completely* understand.
We're integrating that portion of the unit square (in QI with lower left corner at origin) that lies underneath x+y=1. So when 0<z<1, there's only one function that represents the "top": the line x+y=z. And when 1<z<2, there's two functions: y=1 and x+y=z, and therefore we need to break the region of integration into two portions. Got it. Thanks. And, yes. :) 


Register to reply 
Related Discussions  
Moment Generating Functions and Probability Density Functions  Set Theory, Logic, Probability, Statistics  4  
2 different probability density functions will give the same E(g(x)) or not  General Math  1  
Probability density functions  Calculus & Beyond Homework  7  
What's the difference between probability and probability density  Quantum Physics  5  
transformation of nonmonotonic density functions  Calculus & Beyond Homework  0 