# Geometric distribution obtained from exponential distribution

I was trying to solve the following:

Let $\lambda>0$ and $X$ a random variable with $X \sim exp(\lambda)$. Show that $Y=[X]+1$ has geometric distribution of parameter $p=1-e^{-\lambda}$, where $[x]$ denotes the integer part of a number.

I did the following:

$$P(Y=n)=P([X]+1=n)$$$$=P(n-1 \leq X <n)$$$$=F_X(n)-F_X(n-1)$$$$=1-e^{-\lambda n}-(1-e^{-\lambda (n-1)})$$$$=e^{-\lambda (n-1)}-e^{-\lambda n}$$$$=e^{-\lambda (n-1)}(1-e^{-\lambda})$$

If I call $p=1-e^{-\lambda}$, then we have $$P(Y=n)=(1-p)^{n-1}p$$

which is exactly the density function of a geometric random variable with parameter $p=1-e^{-\lambda}$.

I was having doubts with my solution, could anyone tell me if my answer is correct?

– A.S.
Commented Oct 16, 2015 at 5:09
• Clearly explained, well done. Commented Oct 16, 2015 at 6:08

Thanks, the $$p=1-e^{-\lambda}$$ relation was helpful and seems correct. I verified with numpy. Maybe this helps someone.

Note: I replaced $$\lambda$$ by $$-1/\beta$$ ($$beta$$ = scale), since that is what np.random.exponential expects.

I also understand this is not an answer in typical StackExchange fashion, so feel free to delete if it does not belong.

Code:

import matplotlib.pyplot as plt
import numpy as np

N = 10000

for scale in [1,3,7]:
p = 1 - np.exp(-1/scale)

print(f"scale (β) (numpy parameter): {scale}")
print(f"λ: {1/scale}")
print(f"p: {p:.3f}")

rvs1 = np.random.geometric(p=p, size=N)
plt.plot(sorted(rvs1)[::-1])

rvs2 = np.random.exponential(scale=scale, size=N)
plt.plot(sorted(rvs2)[::-1])

plt.xscale("log")


produces

One can see how nicely geometric and exponential line up.