Expected number of frog jumps There is frog jumping forward on a line. Each jump distance is random with a known cumulative distribution function $F$. What is the expected number of jumps to reach (or go beyond) distance $d$ from the origin?

If there is a Wikipedia or other web page describing the formula in terms of $F$, just give a (deep) link. I'd prefer to have the proof as well.
Let $X_i$ denote the distance of jump $i$ of the frog. Each $X_i$ is independent of one-another. The total distance at jump $n$ is then $D_n=X_1+\cdots+X_n$. Define $\tau$ for $D_\tau$, where $\tau$ is the first time such that $D_\tau\geq d$. You are interested in $E(\tau)$. The distribution of $D_n$ is the $n$-fold convolution of $F$'s: $F_n = (F\star)^n:=F\star F\star\cdots \star F$.

$P(\tau&gt; n) = P(D_n&lt;d)$

Now use $E(\tau)=\sum_{n=0}^\infty P(\tau&gt; n)=\sum_{n=0}^\infty P(D_n&lt;d)$

Which we can simplify when $F$ is continuous by noting $P(D_n&lt;d)=P(D_n\leq d)=F_n(d)$. I'm not sure if there's any more simplification you can do without an explicit form for $F$.
The distribution function defines the probability that the random variable is less than or equal to a given x: $P(n\leq x) = F(x)$. In this case, x is the distance of a single jump.

What we need is the expected value V of the distribution function. To get this, we want the derivative of the distribution function, $f(x) = F'(x)$, which produces a function known as the "probability density function" of x. This function will tell us how likely any given jump is to be x units.

Now, given that, the "expected value" of the random variable is equal to:



This "expected value", much like the term states, is the long-term average value of X given a sufficiently large number of trials. For a normal distribution, it is usually the point at which $f(x)$ is at its peak; however many probability functions are not normal (for instance, the probability of an n-sided die rolled once showing a particular face x is 1/n regardless of x). Now, that equation is the general form. We know that the frog always jumps forward, and thus the jump distance is never negative; therefore, we should actually solve the integral for a lower bound of 0:

$$E[x] = \int_0^\infty xf(x)dx$$

$E[x]$ will be the length of the average jump; it has a roughly 50-50 shot to be higher or lower, and thus given enough jumps, the sum of any n jumps should be $E[x] * n$. Now it's arithmetic; solve $E[x] * n &gt;= d$ for n.