I'm reading these notes and would like some help deriving the result at the bottom of page 2.
Suppose that you are building a tower out of unit blocks. Blocks are falling from the sky, [increasing the tower's height one unit at a time]. If you introduce randomness here by declaring the times between arrivals of blocks to be i.i.d. random variables, then you get the simplest 1d random growth model. The kind of question we would like to answer here is what the height $h(t)$ is?
The classical central limit theorem provides the answer: $$ h(t) \approx c_1^{-1}t + \xi c_2c_1^{-3/2}t^{1/2}, $$ where $c_1$ and $c_2$ are the mean and standard deviation of the times between arrivals of the blocks, respectively, and $\xi$ is a standard normal random variable $\mathcal{N}(0, 1)$.
Here's my thought process:
We want to relate the height $h$ and the time $t$ using the central limit theorem, then solve for $h$ in terms of $t$.
Let $X_1, X_2, \cdots$ be the time intervals between the block arrivals. The tower has height $h$ between the times $\sum_{i = 1}^h X_i$ and $\sum_{i=1}^{h+1} X_i$. For simplicity, I'll only use the first sum.
By the central limit theorem, $$ \frac{\sum_{i=1}^h X_i - h \cdot \mathbb{E}[X_1]}{\sqrt{h} \sqrt{\text{Var}(X_1)}} = \frac{t - h \cdot c_1}{\sqrt{h} \cdot c_2} \sim \mathcal{N}(0, 1). $$ So if $\xi \sim \mathcal{N}(0, 1)$, we can write $$ \frac{t - h \cdot c_1}{\sqrt{h} \cdot c_2} \approx \xi, $$ then solve this as a quadratic in $\sqrt{h}$ and square the result.
However, by my calculation, this gives $$ \sqrt{h} \approx \frac{1}{2c_1} \left( - \xi \cdot c_2 + \sqrt{\xi^2c_2^2 + 4tc_1} \right), $$ which is still not correct. (The other root was negative, so I discarded it.)