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The geometric and exponential distributions exhibit memorylessness, i.e., $$ P(X \geq k + j)=P(X\geq k)P(X\geq j) $$

This can be shown for an arbitrary exponential distribution using R:

> pexp(q=1, rate=1, lower.tail = F)
[1] 0.3678794
> pexp(q=0.2, rate=1, lower.tail = F)*pexp(q=0.8, rate=1, lower.tail = F)
[1] 0.3678794

This same result should hold for the geometric distribution. However, when I perform a similar calculation using pgeom(), I do not get the same result

> pgeom(q=3, prob=0.2, lower.tail = F)
[1] 0.4096
> pgeom(q=1, prob=0.2, lower.tail = F)*pgeom(q=2, prob=0.2, lower.tail = F)
[1] 0.32768

Can someone help explain this discrepancy? Based on these examples, the geometric distribution as calculated by pgeom does not exhibit memorylessness.

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    $\begingroup$ Welcome to CV. The memorylessness is for a r.v. $X$ taking values $1$, $2$, $\dots$ and representing the number of trials needed to reach the $1$-st success, while the geometric distribution in R is for the number $Y$ of failures before the first success, with values $0$, $1$, $\dots$. In distribution $X = Y +1$. In R pgeom(k, prob = p, lower.tail = FALSE) gives $P(Y > k) = (1 - p)^{k+1}$ for $k=0$, $1$, $\dots$. $\endgroup$
    – Yves
    Oct 26, 2023 at 9:32
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    $\begingroup$ If you adjust for the support, you will find for example pgeom(q=9-1,prob=0.2,lower.tail=F) and pgeom(q=3-1,prob=0.2,lower.tail=F) * pgeom(q=6-1,prob=0.2,lower.tail=F) give the same 0.2097152 and similarly for other examples $\endgroup$
    – Henry
    Oct 27, 2023 at 0:48
  • $\begingroup$ Ah, I understand. Thank you for the follow-up, Henry. My misunderstanding was a difference in convention. A similar explanation in doi.org/10.1002/9781119536963.ch10 in case helpful for others: "$X$ calculations in R correspond to our $X+1$ calculations above. For example, $X=0$ in R means zero failures before first success, which is equivalent to $X=1$, the first success occurs in the first trial, in our model." Also, in case others use the above commands, the probability is 0.1342177. $\endgroup$ Oct 27, 2023 at 3:53

1 Answer 1

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The memorylessness is for a r.v. $X$ taking values $1$, $2$, $\dots$ and representing the number of trials needed to reach the $1$-st success in a sequence of independent Bernoulli trials, while the geometric distribution in R is for the number $Y$ of failures before the first success, with values $0$, $1$, $\dots$. In distribution $X = Y +1$. In R pgeom(k, prob = p, lower.tail = FALSE) gives $P(Y > k) = (1 - p)^{k+1}$ for $k \geqslant 0$.

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    $\begingroup$ Thanks for the explanation. I wanted to clarify that dgeom in R does not have an argument lower.tail. Also, is there a way to express the property of memorylessness of the geometric distribution in R as I did with the pexp function? $\endgroup$ Oct 26, 2023 at 22:42
  • $\begingroup$ @Henry Thanks for fixing the typo. $\endgroup$
    – Yves
    Oct 27, 2023 at 6:38

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