Friday, September 23, 2016

probability - Quotient Distribution of Positive Independent Random Variables




Suppose X and Y are independent positive random variables with probability density functions fX and fY respectively. Show that Z=X/Y is absolutely continuous and find its probability density function.




My Thought: In order to show that Z is absolutely continuous, I need to show that it has a commulative distribution function. Then I can also differentiate the latter to obtain the probability density function of Z. I have the following calculation
P(X/Yz)=P(XzY,Y<0)+P(XzY,Y>0)=0(yzfX(x)dx)fY(y)dy+0(yzfX(x)dx)fY(y)dy=0(yz0fX(x)dx)fY(y)dy.


Also, if z<0, then we have that FZ(z)=0, since X,Y are positive random variables. Therefore the random variable Z is continuous, and we can differentiate it to obtain the probability distribution function of Z. We have that fZ(z)=0 if z0, and otherwise we have
fZ(z)=0yfX(yz)fY(y)dy.






Is my reasoning above correct?




Any feedback is much appreciated.



Thank you for your time.


Answer



It might be that the PDF cannot be obtained as derivative of the CDF, simply because the CDF is not necessarily differentiable (everywhere).



Fortunately that is not fatal here because there is a more direct route.



For a fixed positive z we find:




P(Zz)=P(XzY)=0zy0fX(x)fY(y)dxdy=0z0fX(uy)yfY(y)dudy=

z00fX(uy)yfY(y)dydu



where the third equality rests on the substitution x=uy.



This shows directly that functiont fZ prescribed by: z0fX(zy)yfY(y)dy

if z>0 and z0 otherwise serves as PDF of positive random variable Z.


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