Suppose we have some i.i.d. sequence $(X\_n)\_{n\in \mathbb{N}}$, and define $Y\_n :=n+X\_n$. What is the probability that $Y\_n$ is increasing from $n$ to $m$?


In more detail, we fix some $n<m \in \mathbb{N}$, and we want to know the probability:
$$\mathbb{P}\{\forall r \in\{n,\dots,m-1\}: Y\_r \leq Y\_{r+1}\} =$$
$$\mathbb{P}\{\forall r \in\{n,\dots,m-1\}: X\_r \leq X\_{r+1} +1\}$$
Now I'd like to use the independence to get an expression in function of $F\_X(t) := \mathbb{P}\{X \leq t\}$ but this isn't possible yet as every $X\_r, r \in \{n+1,\dots,m-2\}$ occurs twice.



Probability that n+X_n is increasing for an i.i.d. sequence
We can compute these probabilities symbolically for some distributions using Mathematica. Letting $k=m-n$, the code is just



```
f[dist_, k_]    := Integrate[(f[dist, k-1] /. w->v) (f[dist, 1] /. u->v) PDF[dist,v],
                             {v, -Infinity, Infinity}]

f[dist_, 1]      = Boole[1 + w > u]

prob[dist_, k_] := Integrate[f[dist, k] PDF[dist, u] PDF[dist,w], 
                             {u, -Infinity, +Infinity}, {w, -Infinity, +Infinity}]

Table[prob[UniformDistribution[{-1, 1}], k], {k, 1, 3}]
Table[prob[ExponentialDistribution[1],   k], {k, 1, 3}]
Table[prob[NormalDistribution[],         k], {k, 1, 2}] //N

```

For the uniform distribution on [-1,1], the first three probabilities are $\frac{7}{8}, \frac{3}{4}, \frac{41}{64}.$ 


For the exponential distribution with parameter 1, the first three probabilities are $\frac{2e-1}{2e}, \frac{6e^3-6e^2+1}{6e^3}, \frac{24e^6-36e^5+6e^4+8e^3-1}{24e^6}.$


For the standard normal distribution, the first three probabilities are 0.760, 0.536 and 0.369, but there were no nice closed forms. For the third normal probability, there was no advantage to the symbolic manipulation, but Mathematica could do a more direct calculation numerically.


