Random variable X is uniformly distributed over the interval [0,1]. $Y = -ln(X)$.I need to find the distribution function for $Y$. So i did this:
$$P(Y \leq x)= P(-ln(X) \leq x) = P(ln(X) \geq -x) = 1 - P(X \leq e^{-x}) = 1 - e^{-x}$$
So distribution function is $1 - e^{-x}$? Is it correct? How to find intervals for it? How to find Expected value for it? Its and integral $\int_0^1 \frac{x}{e^{x}} dx$ but i can't find it
The limits of the integral are now $0$ to $\infty$, because you've changed the density function and it must integrate to 1. Note that $Y$ can now take on values that are between $\lim_{x \rightarrow0}-ln(x) = +\infty$ on the one hand, and $\lim_{x \rightarrow1}-ln(x) = 0$ on the other. So you should rewrite your integral to reflect this, and it will integrate to 1.
$\int_0^\infty e^{-x}dx=1$
The expected value is just $\int_0^\infty x \cdot e^{-x}dx=1$