I have a problem where i am uncertain how to set up a likelihood function

The number of tries before the first success follows a geometric distribution.


In a sample (n=100, iid) a mean number of tries was found to be 7.

So normally without the additional information setting the max. likelihood up as normal.

But now, as i only that that the mean number of tries was seven, not the actual 100 outcomes, can i say that

$$\prod \limits_{n=1}^{100}(1-p)^6*p$$

But solving this i get a negative estimator for p, which doesn’t ring true....

It seems logical to incorporate as much information as possible into my maximum likelihood estimate....

Or could it be possible that i don’t have to integrate the information because there is a point b, (a confidence Intervall) where i would need the mean of x.

  • 1
    $\begingroup$ how do you get a negative estimate? gradient of log likelihood is $100(\frac{1}{p}-\frac{6}{1-p})$. The maximum is at the point $p=\frac{1}{7}$ having 100 identical terms just increases the sharpness of the peak. $\endgroup$ – probabilityislogic May 9 at 5:29
  • $\begingroup$ @probabilityislogic oh i made a stupid mistake, now i get p=1/7, but another question, is the Likelihood function correct? $\endgroup$ – Ang May 9 at 5:33

Your likelihood is correct because, if you have the entire data: $x_1,...,x_{n}$, the likelihood would be $$L(p)=\prod_{i=1}^{n} (1-p)^{x_i-1}p=(1-p)^{\sum x_i -n}p^n$$ which is your expression. Differentiating it (or log of it) will yield $p=1/7$ as your estimate.

  • $\begingroup$ Many thanks, i made a stupid mistake in my calculation! $\endgroup$ – Ang May 9 at 5:35

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