Distribution function exercise Let X be a random variable with density $f_X(x)\frac{x^2}{3}1(−1,2)(x)$
First question
The following scripture :$1(−1,2)(x)$ I interpret in the following way, is right?
$1(−1,2)(x)$ indicates that the function is equal to 1 when $-1 <= x <= 2$ and equal to $0$ otherwise.
Second question which corresponds to the actual exercise: Find the distribution function of $Y = e^X$
For the previous question actually I have the following solution:
For convenience, I call this part of the solution "part one":
We have $P(e^{-1} ≤ Y ≤ e^2) = 1$ and therefore $F_Y(y)=0$ for $y≤e^{-1}$ and $F_Y(y)=1$ for $y≥e^{2}$.
This is the second: For $y ∈ (e^{-1},e^2)$ we have $F_Y(y) = P(e^X ≤ y) = P(X ≤ ln(y)) = \int_{-1}^{ln(y)} \frac{x^2}{3}dx = \frac{1}{3}[\frac{x^3}{3}]_{x=-1}^{x=ln(y)} = \frac{ln(y)^3 - 1(-1)^3}{9} = \frac{(ln(y))^3 + 1}{9}$
I don't understand the need for the reasoning made in the first part, it couldn't be done directly in the second? anyway how do you come to this conclusion? $F_Y(y)=0$ for $y≤e^{-1}$ and $F_Y(y)=1$ for $y≥e^{2}$.
 A: First question: You probably meant to typeset $\mathbf{1}_{(-1, 2)}(x)$. Your definition is almost correct except it should be $-1 < x < 2$ rather than $-1 \le x \le 2$. But it does not really matter for this exercise.


How do you come to this conclusion? $F_Y(y) = 0$ for $y \le e^{-1}$ and $F_Y(y) = 1$ for $y \ge e^2$.

Because $f_X$ is zero outside of the interval $(-1, 2)$ we have $F_Y(y) = P(Y \le y) = P(X \le \ln y) = \int_{-\infty}^{\ln y} 0 \, dx = 0$ if $y \le e^{-1}$.
Similarly, $1-F_Y(y) = P(Y > y) = P(X > \ln y) = \int_{\ln y}^\infty 0 \, dx = 0$ if $y \ge e^2$.


I don't understand the need for the reasoning made in the first part, it couldn't be done directly in the second?

Sure you could, as long as you carefully handle the indicator function. But the casework required to handle the indicator function would basically reduce to the given solution anyway.
$$F_Y(y) = P(Y \le y) = P(X \le \ln y) = \int_{-\infty}^{\ln y} f_X(x) \, dx$$
If $y \le e^{-1}$, then the integral becomes $\int_{-\infty}^{\ln y} 0 \, dx = 0$.
If $y \in (e^{-1} , e^2)$, then the integral becomes $\int_{-\infty}^{-1} 0 \, dx + \int_{-1}^{\ln y} (x^2/3) \, dx = \int_{-1}^{\ln y} (x^2/3) \, dx$.
If $y > e^2$, then the integral becomes $\int_{-\infty}^{-1} 0 \, dx + \int_{-1}^{2} (x^2/3) \, dx + \int_2^{\ln y} 0 \,dx = 0 + 1 + 0 = 1$.
