The problem is:
$Y \sim Unif[0,\theta]$ and we define $X = -Y$.
We want to find $ f_X(x)$ and $F_X(x).$
My solution:
$$F_X(x) = P(X<x)=P(Y>-x)=\int_{-x}^{+\infty}{\frac{1}{\theta} d\theta}$$
But from here I am unsure.I start the problem in this way because if I find the cumulative distribution function, i can derive it to find the Probability Density Function.
Best Answer
The PDF of the continuous uniform random variable Y on the interval $[0,\theta]$ is $f_Y(y)=\frac{1}{\theta}$ for values of $y$ in the given interval.
Therefore, we note that to find the PDF of X, we can do the following:
$F_X(x) = P(X<x)=P(Y>-x)=\int_{-x}^{\theta}{\frac{1}{\theta} dy}$
Note: the reason tht we go up to $\theta$ is that the PDF is $0$ for any value of x outside of the interval $[0,\theta]$
This evaluates to $F_X(x)=1+\frac{x}{\theta}=\frac{x+\theta}{\theta}$ by solving the integral. By differentiating we can find $f_X(x) = \frac{1}{\theta}$ and so we are done as we have found both the PDF and CDF of $X$.
Therefore, $X$~$Uniform[-\theta,0]$ which should be expected if we refer back to the definition of $X$ as being equal to $-Y$.