Shifted Geometric Distribution

Topics: Probability - Distribution Function

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(definition)

Let $X$ be a discrete random variable that handles the repetition of a Bernoulli trial until success is achieved. Let $p$ be the probability of success.

More specifically, let $X$ handle the number of trials needed to get one success (i.e. $R_X=\{1,2,3,\dots \}$).

For such an $X$, we say that it has a shifted geometric distribution and write $X\sim \mathrm{Geo}(p)$.

Non-Shifted Geometric Distribution

When $X$ handles the number of failures before the first success (i.e. $R_X=\{0,1,2,\dots \}$), we say that it has a geometric distribution (non-shifted). See this note for more details on that distribution.

Both of these distributions are commonly simply called geometric, with no other differentiating term. Which of these two distributions is being talked about is a matter of context and/or convention.

There is a convention to refer to the one that counts the trials as the shifted geometric distribution, and that’s what I use in all of my notes.

Density Function §

(theorem)

Such an $X$ has the following density function:

$$f_X(x)=\{\begin{array}{ll}(1-p{)}^{x-1}p& \text{if }x=1,2,3,\dots \\ 0& \text{otherwise}\end{array}$$

Expected Value §

(theorem)

When it comes to the expected value of such an $X$, it is:

$$\mathbb{E}[X]=\frac{1}{p}$$

Variance §

(theorem)

As for the variance of such an $X$:

$$\mathrm{Var}(X)=\frac{1-p}{p^2}$$

Moment-Generating Function §

(theorem)

Finally, such an $X$ has the following moment-generating function:

$$M_X(t)=\frac{p{e}^t}{1-(1-p)e^t}$$