Decreasing Uniform Chain

Step 1: Understand the structure

$X_n$ fails to be the running minimum the first time it's not smaller than everything before it. So right up to that point, every draw was a new record low: the sequence of running minima is strictly decreasing:

$$X_1>X_2>\cdots >X_{N-1},\text{and then}{X}_N>X_{N-1}$$

($N\ge 2$ always: with just one draw, $X_1$ trivially equals $\min \{X_1\}$, so the streak can't break on the first draw.) We want $E[X_{N-1}]$, the value of the running minimum right before it stops being extended.

Step 2: The density of $X_{N-1}$, for a fixed stopping index

Since $N$ itself is random, let's first fix a candidate value $n$ for $N$ and work out the (sub-)density of $X_{n-1}$ landing at some value $x$, conditional on the streak lasting exactly this long. There are three independent requirements packed into this event:

• The $n-2$ earlier draws must all exceed $x$. For $X_{n-1}=x$ to be smaller than everything before it, we need $X_1,\dots ,X_{n-2}>x$. Each is Uniform$(0,1)$ and independent, so this happens with probability $(1-x{)}^{n-2}$.

• Those $n-2$ draws must also land in strictly decreasing order. Being bigger than $x$ isn't enough: they also have to arrive as $X_1>X_2>\cdots >X_{n-2}$ specifically, not in some other order. Since the draws are independent and continuous, all $(n-2)!$ orderings of $X_1,\dots ,X_{n-2}$ are equally likely, so the probability of landing in exactly this one is $\frac{1}{(n-2)!}$.

• The streak must break right after. We need $X_n>x$, which (being an independent Uniform draw) happens with probability $1-x$.

Multiplying these together, and including the density ($=1$ on $(0,1)$) of $X_{n-1}$ itself landing at $x$:

$$f_n(x)=(1-x{)}^{n-2}\cdot \frac{1}{(n-2)!}\cdot (1-x)=\frac{(1-x{)}^{n-1}}{(n-2)!}$$

Step 3: Does this integrate to a valid probability?

Before trusting $f_n$, check that summing it over every possible stopping index $n\ge 2$, and integrating over $x$, gives exactly $1$, since $N$ must take some value.

$${\int }_0^1{f}_n(x)dx=\frac{1}{(n-2)!}{\int }_0^1(1-x{)}^{n-1}dx=\frac{1}{(n-2)!}\cdot \frac{1}{n}=\frac{1}{n(n-2)!}$$

Rewrite this with a common factorial in the denominator: since $n!=n(n-1)(n-2)!$,

$$\frac{1}{n(n-2)!}=\frac{n-1}{n!}=\frac{n}{n!}-\frac{1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$$

This telescopes beautifully when summed from $n=2$ to $\infty$:

$$\sum _{n=2}^{\infty }\left[\frac{1}{(n-1)!}-\frac{1}{n!}\right]=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots =\frac{1}{1!}=1$$

Every term cancels against the next one, leaving just the very first piece. So the total probability really is $1$. This confirms $f_n$ is a legitimate (sub-)density, and we can trust it to compute $E[X_{N-1}]$.

Step 4: Compute $E[X_{N-1}]$

Now weight $x$ by its density and integrate, then sum over $n$:

$$E[X_{N-1}]=\sum _{n=2}^{\infty }{\int }_0^{1}x{f}_n(x)dx=\sum _{n=2}^{\infty }\frac{1}{(n-2)!}{\int }_0^{1}x(1-x{)}^{n-1}dx$$

The inner integral is a Beta integral: ${\int }_0^1{x}^1(1-x{)}^{n-1}dx=\frac{1!(n-1)!}{(n+1)!}$. Substituting in,

$${\int }_0^{1}x{f}_n(x)dx=\frac{1}{(n-2)!}\cdot \frac{(n-1)!}{(n+1)!}=\frac{n-1}{(n+1)!}$$

using $(n-1)!/(n-2)!=n-1$. As in Step 3, split the numerator to make the sum telescope: write $n-1=(n+1)-2$:

$$\frac{n-1}{(n+1)!}=\frac{(n+1)-2}{(n+1)!}=\frac{1}{n!}-\frac{2}{(n+1)!}$$

Step 5: Sum both pieces separately, via $e$

Recall $e={\sum }_{k=0}^{\infty }\frac{1}{k!}$. The first piece, summed from $n=2$:

$$\sum _{n=2}^{\infty }\frac{1}{n!}=e-\frac{1}{0!}-\frac{1}{1!}=e-2$$

The second piece, reindexing $m=n+1$ (so $m$ runs from $3$ to $\infty$):

$$\sum _{n=2}^{\infty }\frac{2}{(n+1)!}=2\sum _{m=3}^{\infty }\frac{1}{m!}=2\left(e-\frac{1}{0!}-\frac{1}{1!}-\frac{1}{2!}\right)=2(e-2.5)$$

Putting it together:

$$E[X_{N-1}]=(e-2)-2(e-2.5)=e-2-2e+5=3-e$$

Step 6: Final answer

We're told $E[X_{N-1}]=a+be$, so matching $3-e$ gives $a=3$, $b=-1$, hence

$$a+b=3+(-1)=2$$

As a quick sanity check, $3-e\approx 0.282$, comfortably inside $(0,1)$ as it must be, and noticeably smaller than $0.5$, which makes sense: $X_{N-1}$ is, by construction, the bottom of a streak of record lows, so it's biased well below the average of a single Uniform$(0,1)$ draw.

Thus, $a+b=2$