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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?

Bryden C
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    By the way, here are some approaches to computing the PDF of the range: https://math.stackexchange.com/questions/33767/density-and-expectation-of-the-range-of-a-sample-of-uniform-random-variables. – Minus One-Twelfth Feb 18 '19 at 00:43
  • The first attempt is only partially correct because you are using only the marginals. The min $Y_{(1)}$ and max $Y_{(n)}$ are correlated with the joint shown in the answer (by user4143) of the linked post as commented by Minus One-Twelfth. The 2nd attempt is also only partially correct because you are dealing with conditional probability while not being aware of it. Again, see the answer by Chad in the same post. – Lee David Chung Lin Feb 20 '19 at 14:17
  • Does this answer your question? Distribution of range of Uniform $(0,1)$ distribution – StubbornAtom Dec 05 '21 at 14:10

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