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Show that: $$P(N(x)>n)= \frac{x^n}{n!}$$ I was thinking that we can take $X_1+...+X_n$ to be a single random random variable $S_n$ and that the $n$ in the $P(N(x)>n)$ is not the same as the $n$ in the definition of $N(x$).
But unsure how to actually start proceeding to prove this.

Any help appreciated, thank you.

2 Answers2

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You may prove this by induction. For $n=1$, the equality clearly holds, i.e., $$ \mathsf{P}(N(x)>1)=\mathsf{P}(S_1\le x)=x. $$ For the inductive step note that \begin{align} \mathsf{P}(N(x)>n)&=\mathsf{P}(S_n\le x)=\mathsf{E}[\mathsf{P}(S_{n-1}\le x-X_n\mid X_n)] \\ &=\int_0^1 \mathsf{P}(S_{n-1}\le x-y)\,dy =\int_0^x \frac{(x-y)^{n-1}}{(n-1)!}\,dy=\frac{x^n}{n!}. \end{align}

Using this result, one may compute the probability generating function by noticing that $$ \sum_{i=0}^\infty\frac{(tx)^{i}}{i!}=e^{tx}. $$ That is, $$ \mathsf{E}t^{N(x)}=\sum_{i=0}^\infty t^i\mathsf{P}(N(x)=i)=t\sum_{i=1}^\infty \frac{(tx)^{i-1}}{(i-1)!}-\sum_{i=1}^\infty \frac{(tx)^{i}}{i!}=te^{tx}-(e^{tx}-1). $$

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Not as slick at d.k.o.'s approach but a little more direct. Assume $n>1$.

Notice for $k=n+1, n+2, \ldots$ we have $$P(N(x)=k)=P(S_{k}>x,S_{k-1} \leq x)=P(S_{k-1} \leq x,X_k>x-S_{k-1})$$

Since $S_{k-1}$ and $X_k$ are independent, we can say

$$P(N(x)=k)=\int_0^xP(X_k>x-t)f_{S_{k-1}}(t)dt=\int_0^x\frac{t^{k-2}}{(k-2)!}(t+1-x)dt$$

The last equality utilized the Irwin Hall distribution for the sum of $n$ iid uniformly distributed random variables. Evaluating this integral gives $$P(N(x)=k)=\frac{x^{k-1}}{(k-1)!}-\frac{x^k}{k!}$$

So finally,

$$P(N(x)>n)=P(N(x)=n+1)+P(N(x)=n+2)+\ldots=\sum_{k=n+1}^{\infty}\bigg[ \frac{x^{k-1}}{(k-1)!}-\frac{x^k}{k!} \bigg]=\frac{x^n}{n!}$$