cumulative distribution functiondistributionsprobabilityuniform distribution
(from The Probability Tutoring Book, C. Ash, p. 157)
Find the density of $Y$ if $Y = 1/X$ and $X$ is uniform on $[-1,1]$.
The distribution function given in the answer key is
$$
F(y) =
\begin{cases}
\frac{-1}{2y} & \text{if }y \leq -1\\
\frac{1}{2} & \text{ if }-1 \leq y \leq 1\\
1-\frac{1}{2y} & \text{if } y \geq 1
\end{cases}
$$
My question is why the first and last cases aren't $0$ and $1$. Doesn't $Y$ only range over $[-1, 0)$ and $(0, 1]$?
Here is a hint.
Consider carefully the term $\mathbb P( X \leq z Y \mid z Y < 1 )$. In particular, for concreteness, choose $z = 2$, so that we are considering the event $\mathbb P( X \leq 2 Y \mid Y < 1/2 )$.
Now, look at this picture (which is very closely related to the above probability).
Now, does that conditional probability depend on our particular choice of $z$?
You seem to be a little confused about the likelihood function of the $U[0,\theta]$ model.
Let $f(x\mid\theta)=1/\theta$, for $0\leq x\leq\theta$, and $f(x\mid\theta)=0$, otherwise, for $\theta>0$.
For some fixed $x$, what is the graph of $f(x\mid\theta)$ as a function of $\theta$?
To draw the graph, notice -- and this is the key point -- that $x\in[0,\theta]$ if and only if $\theta\in[x,\infty)$.
So, using indicator functions, we have $$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty)}(\theta)\, .$$
After you understand this, just do the integration: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$
Best Answer
I hope a quick simulation in R will help you visualize this transformation. For readable graphs, I use only transformed values in $(-10,10).$
Note: The ECDF of a sample plots values of the sample from smallest to largest. It is a stairstep function (with jumps too small to see here). At each value of a sample of size $n$ the ECDF jumps up by $1/n.$ If there are $K$ values tied at a point, then the function jumps by $k/n$ at that point. (No ties here) For a sufficiently large sample the ECDF closely imitates the population CDF.