Uniform distribution transform: e^2x

[TOOP]

X is a uniformly distributed random variable on a [-1, 1] range. (i.e. X is U(-1, 1))
Find the distribution of e^2X:

I feel like it has something to do with the uniform's relation to exponential function,
but i get stuck.

I begin by using inverse transform:
Fy(y) = Fx[ln(y)/2]
fy(y) = fx[ln(y)/2]*[1/(2y)]
fy(y) = 1/(4y)

please help

 

[lanedance]

so you have y(x)=e^(2x)

the infinitesimal increment dy is related to the increment dx by dy = 2e^(2x)dx

say the probability to be in the increment dx is p(x)dx, with probability distribution p(x)

clearly the probability to be in the related increment dy is the same, so let's call the probability distribution of y, q(y), then we have

q(y)dy = p(x)dx

can you take it from here?

 

[Mute]

X is a uniformly distributed random variable on a [-1, 1] range. (i.e. X is U(-1, 1))
Find the distribution of e^2X:

I feel like it has something to do with the uniform's relation to exponential function,
but i get stuck.

I begin by using inverse transform:
Fy(y) = Fx[ln(y)/2]
fy(y) = fx[ln(y)/2]*[1/(2y)]
fy(y) = 1/(4y)

please help

What exactly do you need help with? It looks like all you have left to do is give the range of y, which should be simple: if X can have values between -1 and +1, what is the range of values of y? Once you know the range, you can check that your result works by checking that

$$\int dy \frac{1}{4y} = 1,$$

where the integral is over the range of y that you need to determine.

 

[lanedance]

not quite, the OP needs to find the distribution function of the random variable Y, based on its relation to the random variable X which is distributed constantly over the region (-1,1)

EDIT: i haven't actually whether the OPs distribution was correct as I didn't follow the steps ;(, however i think i see what's going on now

 

[Mute]

not quite, the OP needs to find the distribution function of the random variable Y, based on its relation to the random variable X which is distributed constantly over the region (-1,1)

EDIT: i haven't actually whether the OPs distribution was correct as I didn't follow the steps ;(, however i think i see what's going on now

Your method will indeed give the same result as the OP's method. (really, they're the same method, just dressed up a little differently)