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Because I've read it is either

  • $g(x) = \prod_{i=1}^∞\ p(1-p)^{x_i}$

or

  • $g(x) = \prod_{i=1}^∞\ p(1-p)^{x_i-1}$

So I'm really confused.

Reference: https://math.stackexchange.com/questions/4429910/mle-of-the-geometric-distribution

Edit: My iid observations go from $x_1, x_2$ etc ...

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  • $\begingroup$ Please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Nov 8 at 13:34
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    $\begingroup$ Hint: stats.stackexchange.com/questions/560497/… $\endgroup$
    – kjetil b halvorsen ♦
    Commented Nov 8 at 13:41
  • $\begingroup$ Distributions don't have likelihood functions. Are you perhaps referring to the probability function? $\endgroup$
    – whuber ♦
    Commented Nov 8 at 13:43
  • $\begingroup$ There are two different versions of the geometric in common use. The number of trials version and the number of failures version. Which do you need? $\endgroup$
    – Glen_b
    Commented Nov 9 at 0:22

1 Answer 1

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The geometric distribution is counting the number of trials/failures in a series of Bernoulli trials untill a success occurs.

The difference between $x_i$ and $x_i-1$, is whether $x_i$ refers to the number of trials, or the number of failures (which differ by one).

Btw, likelihood functions are typically described as functions of the parameter: $g(p)$, instead of $g(x)$.

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