Question

F(x)= 0 for x<0, x for 0=1. Hence find the cumulative distribution function of the random variable Y =-2lnX.

Answer 1

I have yo wonder about this ...

i think it might be the improper integral:

\[\int_{0}^{1}\ -2ln(x)\ dx\]

but I aint sure

Answer 2

Hey, thanks for your reply, can you show me your reasoning behind the parameters 0 & 1? The irritating thing is I can do this question if I work out the parametes, I just don't know what they are.

Answer 3

i was going on a gut feeling, but after reading up on it i see its wrong for 0 to 1; the cumulative is defined for everything between certain zscores; and the basic CDF is from -inf to z

Answer 4

this looks like its going to have to be set up as a piecewise finction

Answer 5

http://www.math.vt.edu/people/qlfang/class_home/Lesson2021.pdf

Answer 6

Yeah I've seen that, I've been researching this for about 5 hours now and have not found a single thing that either explains how to calculate the c.d.f without the p.d.f or that tells you how to work out a and b. I've also found zero examples using a log function.

Answer 7

the pdf is defined by the piecewise at the start; you can draw up a pdf from that i believe ...

Answer 8

F(x)= 0 for x<0, x for 0<x<1, 1 for x>=1

Answer 9

another view from might be this:



but i gotta read a bit more to see what to do with these

Answer 10

the pdf is the derivative of the cdf; which is why we can integrate up into it

Answer 11

i know that ln(x) is undefined for values less than or equal to zero; and for a value of "1" it equals zero; so the only viable interval that makes sense to me is between 0 and 1.

but again, I have no proof for that yet

Answer 12

amistre please help meee

Answer 13

"We'll learn how to find the probability density function of Y, using two different techniques, namely the distribution function technique and the change-of-variable technique. "

https://onlinecourses.science.psu.edu/stat414/book/export/html/128

Answer 14

if im reading this right, then since Y is increasing, we can use the DFT;

\[F_y(y)=P(Y\le y)=P(-2ln(X)\le y)\]
\[=P(-ln(X)\le \frac{1}{2}y)\]
\[=P(ln(1/X)\le \frac{1}{2}y)\]
\[=P(e^{ln(1/X)\le \frac{1}{2}y})\]
\[=P(1/X\le e^{\frac{1}{2}y})\]
\[=P(X\le e^{-\frac{1}{2}y})\]

hmmm

Answer 15

forgot to switch over my inequals sign ...

Answer 16

\[F_y(y)=P(X\ge e^{-\frac{1}{2}y})\]
might be better

Answer 17

opps, I forgot that -2ln(X) is the flipped version of ln .... so its decreasing; gotta read up a little more :)

Answer 18

same steps in the decreasing except for its the other side; so we use that in the 1-Fx version

Answer 19

1) if its an increasing function for Y = uX, then we can find the Fx

2) if its decreasing, we go with the 1-Fx ...

Answer 20

so, given this peicewise Fx stuff, lets see what we can do:

F(x)= 0 for x<0, x for 0<x<1, 1 for x>=1

\[F(x)=0;\ x<0\]
\[\Large 1-\int_{}^{e^{-\frac{1}{2}y}}0dx\]

hmmm, just gotta see how they are getting the interval for integration ....

Answer 21

Thanks a lot for all your help, I get it now :).

Answer 22

ok, the interval for x on this is from (-inf,0) so the range is what we use for our integration interval since we are ingrating across the "y"


whew!! .... im glad you do :) cause my workings here are a bit iffy to me