Question

is this right?

Answer 1

consider \[F_x(x)=1-e^{-3x}\] when x>0
let \[y=e^x\]find pdf of y

Answer 2

this is what i did

Answer 3

\[F_y(y)=p(Y \le y)\]\[=P(e^x \le y)\]\[=P(x \le \ln(y))\]\[=F_x(\ln(y))=1-e^{-3\ln(y)}=1-y^{-3}\]
so pdf of y
\[f_y(y)=\frac{dF_y(y)}{dy}=3y^{-4}\]

Answer 4

lol thanks for trying @FeoNeo33

Answer 5

lol i just saw ur post ... i saw that guys post when i was typing sorry

Thanks for trying @lgbasallote :D:D

Answer 6

no thanks?*

Answer 7

Yay :p

Answer 8

hehe

Answer 9

I'm not familiar with this notation - you may to explain a bit before I can attempt to help.

Answer 10

pdf of y?

Answer 11

\[F_y(y)=p(Y \le y)\] ?

Answer 12

FFM just let me know that pdf is Probability Distribution Function. I haven't really studied those.
Sorry.
<-- only an aeronautical engineer :)

Answer 13

oh lol i was expalaining,..
and thanks for viewing it :D i appreciate it @FoolForMath @asnaseer

Answer 14

yw - sorry for not being able to help though...