# Relation between Geometric and Exponential random variables

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.

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 $$\theta = -\frac{\log(1-p)}{\lambda}.$$