Distribution of minimum of difference between uniforms $(0,1)$ 
Let $U_1,...,U_N$ be a sequence of independent uniformly $(0,1)$ random variables. Define $Y_k := U_k - U_{k+1}$. Find the distribution of $\min\left\{Y_k\right\}_{k=1}^{N-1}$.

I've get the distribution of $U_k - U_{k+1}$, but I wasn't able to get the distribution of minimum. Any ideas? 
 A: I will give my working for the case $N = 2$ which is already quite tricky, and will have another attempt to think about the general $N$ case. But hopefully this might 
help! 
Let $U_0,U_1,U_2$ be three random variables uniform on [0,1] and define $Y_i = U_i - U_{i+1}$ for $i=1,2$. Define $Z = \min(Y_0,Y_1)$ (apologies for changing your indexing - it's just how I've calculated it and I don't want to confuse myself when typing up!).
By conditioning on the value of $U_1$ we get:
$$ \begin{align}1 - F_Z(z) = P(Z \geq z) & = P(Y_0 \geq z, Y_1 \geq z) \\
& = \int_0^1P(Y_0\geq z,Y_1\geq z\,|\,U_1 = x)f_{U_1}(x)dx \\
& = \int_0^1P(Y_0\geq z,Y_1\geq z\,|\,U_1 = x)dx\end{align}$$
So we are interested in $P(Y_0\geq z,Y_1\geq z\,|\,U_1 = x)$. Now,
$$ \begin{align} P(Y_0\geq z,Y_1\geq z\,|\,U_1 = x) & = P(U_0 - x \geq z) \times P(x - U_2 \geq z) \\ & = P(U_0 \geq x+ z) \times P(U_2 \leq x-z)\end{align}$$
We have to be a bit careful about $P(U_0 \geq x+z)$ and $P(U_2 \leq x-z)$ as $z$ can be negative. We find 
$$ P(U_0 \geq x + z) = \left\{ \begin{align} & 1 & z+x\leq 0 \\ & 1-(x+z) & 0 \leq z + x \leq 1 \\  & 0 & z + x \geq 1\end{align} \right. $$
and 
$$ P(U_2 \leq x - z) = \left\{ \begin{align} & 0 & x-z\leq 0 \\ & x-z & 0 \leq x-z \leq 1 \\  & 1 & x-z \geq 1\end{align} \right. $$
Having a look at the quick sketch at the bottom, we can see that the integration we need to do will be different for each of the regions $-1 \leq z \leq -\frac{1}{2}$,$-\frac{1}{2} \leq z \leq 0$, $0 \leq z \leq \frac{1}{2}$, $\frac{1}{2} \leq z \leq 1$. Let's do them one by one.
For $-1 \leq z \leq -\frac{1}{2}$:
$$ \begin{align} \int_0^1P(Y_0\geq z,Y_1\geq z\,|\,U_1 = x)dx & = \int_0^{1+z} 1-z-x \;dx + \int_{1+z}^{-z} dx + \int_{-z}^1 x-z\;dx \\
& = -2z^2 - 4z + 1 \end{align}$$
For $-\frac{1}{2} \leq z \leq 0$ : 
$$ \begin{align} \int_0^1P(Y_0\geq z,Y_1\geq z\,|\,U_1 = x)dx & = \int_0^{-z} 1-z-x \;dx + \int_{-z}^{1+z} (x-z)(1-x-z)\;dx \\ & \qquad + \int_{1+z}^1 x-z\;dx \\
& = \frac{4}{3}z^3 - 3z^2 - 3z + \frac{2}{3} \end{align}$$
For $0 \leq z \leq \frac{1}{2}$ : 
$$ \begin{align} \int_0^1P(Y_0\geq z,Y_1\geq z\,|\,U_1 = x)dx & = \int^{1-z}_z (x-z)(1-x-z)\;dx \\ & = -2z^3 + 2z^2 - z + \frac{1}{6} \end{align} $$
and finally for $\frac{1}{2} \leq z \leq 1$ : 
$$ \int_0^1P(Y_0\geq z,Y_1\geq z\,|\,U_1 = x)dx = 0$$
Putting all those threads together should hopefully give you some sort of answer (which at least passes sanity checks at $z=-1$ and $z=1$ but otherwise may be completely wrong). Hopefully even if completely wrong, this partial attempt could inspire you to the correct answer.

