i need some help to understand what is going on here.I'll be more specific on questions after i give the context.
Suppose we want to simulate $X$ such that:
$P\{X = i \} = p(1-p)^{i-1} , i \geq 1 $
As:
$\sum\limits_{i=1}^{j-1} P\{X = i\} = 1- P\{X > j-1 \} = 1-(1-p)^{j-1}$
we can simulate such a random variable by generating a random number $U$ and setting $X$ equal to that value $j$ for which:
$1-(1-p)^{j-1} < U < (1-p)^j$
or,equivalently, for which:
$(1-p)^{j} < 1-U < (1-p)^{j-1}$
As, $1-U$ has the same distribution as $U$ , we an thus define $X$ by :
$X = min\{j:(1-p)^j < U \} = min \big \{j:j > \frac{log (U)}{log(1-p)} \big\} = 1 +\big[ \frac{log (U)}{log(1-p)}\big]$
questions:
$\textbf{(1)}$ $\sum\limits_{i=1}^{j-1} P\{X = i\} = 1- P\{X > j-1 \} = 1-(1-p)^{j-1}$ this is nothing more than specifying the CDF (distribution function) right?
$\textbf{2)}$ $1-(1-p)^{j-1} < U < (1-p)^j$ . X gets the value of $j$ if the random number $U$ is between $1-(1-p)^{j-1}$ and $(1-p)^j$ .Did i get it right?
$\textbf{3)}$
$X = min\{j:(1-p)^j < U \} = min \big \{j:j > \frac{log (U)}{log(1-p)} \big\} = 1 +\big[ \frac{log (U)}{log(1-p)}\big]$
can someone show me or explain the stepwise calculations so that you finally come to $1 +\big[ \frac{log (U)}{log(1-p)}\big]$ , because u don't get the given calculations(in the way they do it)
(1) Yes, this is exactly the CDF.
(2) Yes, this is exactly what this means.
(3) This is a simple computation: $$\begin{align*} (1-p)^j<U &\Leftrightarrow \log(1-p)^j<\log U\\ &\Leftrightarrow j\log(1-p)<\log U\\ &\Leftrightarrow j>\frac{\log U}{\log(1-p)}\\ \end{align*}$$ Note that we have used two facts: $\log$ is monotone increasing, and therefore it preserves inequalities, and $\log(1-p)$ is negative, and therefore it reverses the inequality when we divide by it.
Finally, we note that the smallest integer $j$ such that $j>\frac{\log U}{\log(1-p)}$ is, by definition, the ceiling of $\frac{\log U}{\log(1-p)}$, which is also equal to $1+\lfloor \frac{\log U}{\log(1-p)}\rfloor$.