Let $Y_1, Y_2,..., Y_n$ denote a random sample from the uniform distribution $f (y) = 1, 0 ≤ y ≤ 1.$ Find the probability density function for the range $R = Y_{(n)} − Y_{(1)}$.
First Attempt:
Since $F(y_{(i)}) = y_i,$ for $ 0 \leq y_i \leq 1$, we have
$$ F(y_{(1)}) = 1-(1-y)^n \qquad \qquad F(y_{(n)}) = y^n $$
Thus,
$$ f(y_{(1)}) = n(1-y)^{n-1} \qquad \qquad f(y_{(n)}) = ny^{n-1} $$
From here, I argued that $Y_{(n)} = R + Y_{(1)}$, and I tried (unsuccessfully) to do a transformation that allowed me to find the distribution of R. Is this method possible?
Second attempt:
Force $Y_{(1)} = a$. Then $P(a \leq R \leq b)$ is the probability that the other $n-1$ observations fall withing the interval $[a,b]$, or thus the range is $r = b-a$. Since $Y$ is uniform,
$$ P(a \leq R \leq b) = (b-a)^{n-1} = r^{n-1} $$ $$ f(r) = \frac{d}{dr} r^{n-1} = (n-1)r^{n-2} $$
which is the pdf of the range.
Is this answer correct? Further, is there a way to finish my first attempt to get this same answer?