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Let $Y_n$ be a random variable following the exponential distribution with $\lambda$ as a parameter, $Y_n \sim \mathcal E (\lambda)$

For $\theta > 0$, we define $X_n$ as following :

$$X_n = [\theta Yn]$$ Where $[\theta Y_n]$ denotes the superior whole part :

$$[x]=min\{k\in \mathbb Z, k\geq x\}$$

How to find $\theta$ such that $X_n$ follows a geometric distribution ? $X_n \sim \mathcal G (p)$ ?

$\textbf{My attempt : }$ \begin{align} \mathbb P (X_n=j) &=\mathbb P(min\{k\in \mathbb Z, \theta Y_n \leq k\}=j)\\ &=\mathbb P(\theta Y_n\leq j)\\ &=\mathbb P(Y_n\leq \frac{j}{\theta}) \end{align} Since $Y_n \sim \mathcal E (\lambda)$, then: $$\large \mathbb P(Y_n\leq \frac{j}{\theta})=1-e^{-\lambda\frac{j}{\theta}}$$ And that's where I got. I didn't quite understant the part of the min.

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You define $$[x] = \min\{k \in \mathbb Z, k \ge x\}.$$ This means, $[x]$ is the smallest integer $k$ such that $k \ge x$. So for instance, $[3.2] = 4$, because $3 < 3.2 < 4$.

In the mathematical literature, what you define as $[x]$ is more commonly written $\lceil x \rceil$, called the ceiling of $x$. I will use this notation but the meaning is the same.

What does it mean when we say $$X_n = \lceil \theta Y_n \rceil?$$ Specifically, what is the relationship between $\Pr[X_n = x]$, the probability that $X_n = x$, and the corresponding values of $Y_n$? By definition, $$\Pr[X_n = x] = \Pr[\lceil \theta Y_n \rceil = x].$$ But what values of $Y_n$ satisfy $\lceil \theta Y_n \rceil = x$? For example suppose we are interested in the case $x = 4$. Then $\theta Y_n$ must be strictly greater than $3$, but less than or equal to $4$. Since if $\theta Y_n = 3$, then $\lceil \theta Y_n \rceil = 3$; and if $\theta Y_n = 4.00001$, then $\lceil \theta Y_n \rceil = 5$--we still have to "round up" to the nearest larger integer. So $$\Pr[\lceil \theta Y_n \rceil = x] = \Pr[x-1 < \theta Y_n \le x], \tag{1}$$ where we remind ourselves that $x$ must be an integer.

Now, we can apply the fact that $Y_n$ is exponential with rate $\lambda$: $$\begin{align} \Pr[x - 1 < \theta Y_n \le x] &= \Pr\left[\frac{x-1}{\theta} < Y_n \le \frac{x}{\theta}\right] \\ &= F_{Y_n}(x/\theta) - F_{Y_n}((x-1)/\theta) \\ &= (1 - e^{-\lambda x/\theta}) - (1 - e^{-\lambda (x-1)/\theta}) \\ &= e^{-\lambda (x-1)/\theta} - e^{-\lambda x/\theta} \\ &= (1 - e^{-\lambda/\theta}) e^{-\lambda (x-1)/\theta}. \tag{2} \end{align}$$

where $F_{Y_n}(y) = 1 - e^{-\lambda y}$ is the cumulative distribution function for an exponential random variable with rate $\lambda$. Notice that we've written the probability as two factors, the first of which is independent of $x$, depending only on fixed constants $\lambda$ and $\theta$.

What does a geometric distribution look like? Well, suppose $W$ is geometric with parameter $p$; e.g., $$\Pr[W = w] = (1-p)^{w-1} p, \quad w \in \{1, 2, 3, \ldots \}. \tag{3}$$ Notice that I have chosen a parametrization such that $W$ has positive integer support; some parametrizations include $0$, but because $\lceil x \rceil$ for $x > 0$ is never $0$, we won't use such a parametrization.

Then all that is left is to compare the geometric PMF in Equation $(3)$ against the result we got in Equation $(2)$. Here, $x$ and $w$ must play the same role. What is $p$? Clearly, we must have $$(1-p)^{w-1} = e^{-\lambda (x-1)/\theta} = (e^{-\lambda/\theta})^{x-1},$$ so $$p = 1 - e^{-\lambda/\theta},$$ and we write $$\Pr[X_n = x] = (1-p)^{x-1} p, \quad p = 1 - e^{-\lambda/\theta}, \quad x \in \{1, 2, 3, \ldots\}.$$

What this tells us is that for any $\theta > 0$, the distribution of $\lceil \theta Y_n\rceil$ is geometric with parameter $p = 1 - e^{-\lambda/\theta}$. If you want a specific $p$, then the value of $\theta$ that gives you a geometric distribution with such a $p$ is $$\theta = -\frac{\log(1-p)}{\lambda}.$$

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