# Uniform distribution transform: e^2x

1. Sep 15, 2011

### 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)

2. Sep 15, 2011

### 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 probabilty distribution p(x)

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

q(y)dy = p(x)dx

can you take it from here?

3. Sep 15, 2011

### Mute

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.

4. Sep 15, 2011

### 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 whats going on now

5. Sep 15, 2011

### Mute

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)