This is a question I have to solve and need help with. I know it's usual to give pointers and hints so the OP can follow from there. Thus, I'll appreciate all input that shows me the way to go.

Let $X$ be a non-negative random variable. Let $Y = ln(X)$.

Let $f_{X}(x)$ be:

$ f_{X}(x)=\begin{cases} 1/4, & \text{if 2 < $x$ $\leq6$ }\\ 0, & \text{otherwise}. \end{cases}$

And let $f_{Y}(y)$ be:

$ f_{Y}(y)=\begin{cases} g(y), & \text{if a < $x$ $\leq b$ }\\ 0, & \text{otherwise}. \end{cases}$

What is the formula for $g(y)$? What are the values of $a$ and $b$?

I know that

$F_{Y}(y) = \mathbb{P}(Y \leq y)$

$F_{Y}(y) = \mathbb{P}(ln(X) \leq y)$

$F_{Y}(y) = \mathbb{P}(X \leq e^y)$

$F_{Y}(y) = F_{X}(e^y)$

From this point on, I am not sure what I should plug in to find $g(y)$.

Should I have converted the range of $X$, i.e. $[2,6]$, to a range for $Y$? The histogram for the variable $X$, as seen in the figure below, was obtained with the sampling of 1 million data points from the distribution $U:[2,6]$. The histogram for $Y$ followed after taking the $ln(.)$ of $X$. The range of $Y$ is $[ln(2), ln(6)]$.

Histograms for both X and Y = ln(X)


Knowing that the pdf is the derivative of the cdf, from $F_Y(y) = F_X(e^y)$, to get the pdf of $Y$ you just need to differentiate with respect to $y$ and you get that $f_Y(y) = f_X(e^y)e^y$. So if $y\in[ln(2),ln(6)]$ it is $e^y/4$ and $0$ otherwise.


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