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 $


$\sum\limits_{i=1}^{j-1} P\{X = i\} = 1- P\{X > j-1 \} = 1-(1-p)^{j-1}$

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 :

$X = min\{j:(1-p)^j < U \} = min \big \{j:j > \frac{log (U)}{log(1-p)} \big\} = 1 +\big[ \frac{log (U)}{log(1-p)}\big]$


$\textbf{(1)}$ $\sum\limits_{i=1}^{j-1} P\{X = i\} = 1- P\{X > j-1 \} = 1-(1-p)^{j-1}$ 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?


$X = min\{j:(1-p)^j < U \} = min \big \{j:j > \frac{log (U)}{log(1-p)} \big\} = 1 +\big[ \frac{log (U)}{log(1-p)}\big]$

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 $1+\lfloor \frac{\log U}{\log(1-p)}\rfloor$.


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