Transformation of the uniform distribution

[stukbv]

Homework Statement

I am told that X is a random variable with uniform distribution over [0,1]
I need to find the mean and variance of log(X)

2. The attempt at a solution
I assume I must find the pdf of log(X) so I did this as follows;

Let Y=log(X)
Then to find the cumulative distribution function I considered;
P(Y≤ y) = P(log(X)≤ y) = P( X < e^y)

I know that X is uniform over [0,1] so this is equal to
∫1.dy between e^y and 0, where e^y≤ 1

This gives me that the cumulative distribution function = e^y
Which tells me that the probability density function is also e^y

Now what I am not sure of is what must y lie between, is this for y between e^b and e^a ?

 

[Ray Vickson]

Homework Statement

I am told that X is a random variable with uniform distribution over [0,1]
I need to find the mean and variance of log(X)

2. The attempt at a solution
I assume I must find the pdf of log(X) so I did this as follows;

Let Y=log(X)
Then to find the cumulative distribution function I considered;
P(Y≤ y) = P(log(X)≤ y) = P( X < e^y)

I know that X is uniform over [0,1] so this is equal to
∫1.dy between e^y and 0, where e^y≤ 1

This gives me that the cumulative distribution function = e^y
Which tells me that the probability density function is also e^y

Now what I am not sure of is what must y lie between, is this for y between e^b and e^a ?

For X between 0 and 1, what is the range of log(X)?

Anyway, to get the mean and variance of Y = log(X), you don't need the distribution of Y, although getting it is certainly one way of doing the problem.

RGV

 

[stukbv]

I think my lecturer wants me to do it this way. So y must be between - infinity and 0?
Is my pdf correct now?