Obviously I'm stuck. The question is to find the density function, Y = 2X.
Fy(a) = Pr{Y<=a}
= Pr{2X<=a}
= Pr{X<=a/2}
= Fx(a/2)
Differentiate both Fy and Fx cdfs
fy(a) = 1/2fx(a/2)
I don't understand where the 1/2 comes from. Differentiating a cdf like F(a) results in the density function f(a), so wouldn't differentiating Fx(a/2) produce fx(a/2)?
Thanks! I got it now.
I'm also looking for the distribution function of Y with Y=e
F_y(x) = P{Y<=x}
= P{e <= x}
= P{X <= log(x)}
= f(y)d(y)
= log(x)
I'm getting stuck at the integral stage. Could you explain how to get from the integral to the answer of log(x)? Isn't it just f(log(x))-0 = f(log(x))?
Ok. I have a density function for X: f(x) = . The question asks me to find E[e ].
To simplify, Y = e . To determine Fy (cdf of Y), for
Fy = Pr(Y x)
= Pr(e )
= Pr( )
= f(y)dy **My query is at this stage**
= log(x)
Then, differentiate Fy so that fy = (when )
After a bit more integration the answer comes to e - 1. I don't get how the integral marked with an asterisk is calculated though.
I see you didn't take my advice regarding notation - thta's probably part of your trouble.
for .
(Note that ).
Then the pdf for Y is for and zero elsewhere.
Then .
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But why do all this when you can just use where f(x) is the pdf of X ......!?
In your case . Much less effort!
Perfect. That's just what I wanted. Thanks!
I was just trying to understand the derivation. The notation that I used was as in the probability book I'm using - perhaps I should dump it? I just copied out the notation there in case I made a mistake. But I agree, it's simpler the way you put it there. cheers!