# Fitting to data with a Bernoulli (I think) distribution

I have a series of data to which I want to fit my model. The model predicts the probability of success at a given value of x. I have a single data point at a number of points in this space. As I have a single point which is either pass or fail with a certain probability I believe I should fit using a Bernoulli maximum likelihood fit, is this correct? So i have a likelihood function which looks something like $$L(\theta,x) = \Pi^{n}_{i} \theta^{x_i}(1-\theta)^{1-x_i}$$ where n are my data points? This is just different enough to the usual Binomial likelihood case to have completely thrown me.

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Isn't $\prod_i^n\theta^{x_i}(1-\theta)^{1-x_i}$ the same as $\theta^{\sum_i x_i}(1-\theta)^{n-\sum_i x_i} = \theta^y(1-\theta)^{n-y}$ where $y = \sum_i x_i$ is a binomial $(n,\theta)$ random variable? –  Dilip Sarwate Dec 21 '11 at 17:02
Wouldn't $y$ have the distribution ${n \choose k} \theta^y (1 - \theta)^{n-y}$ if $y = \sum_{i=1}^n x_i$ –  Sacha Epskamp Dec 21 '11 at 17:53
@SachaEpskamp What is the relationship between $y$ and $k$? In any case, the probability mass function of $y$ is irrelevant. The likelihood of the observation or data $$x = (x_1,x_2,\ldots,x_n)$$ is exactly what Bowler stated it to be, and it is essentially a binomial likelihood. The extra constant factor $\binom{n}{y}$ (remember that $y$ is a function of the data $x$ and is thus fixed) that you have included makes no difference: the value of $\theta$ that maximizes $\theta^y(1-\theta)^{n-y}$ is the same as that which maximizes $\binom{n}{y}\theta^y(1-\theta)^{n-y}$. –  Dilip Sarwate Dec 21 '11 at 22:12

If $x$ is a vector with independent measurements of a single variable which can be pass or fail, then this is the likelihood indeed, except that I think $L(\boldsymbol{x} \mid \theta)$ would be the proper notation.

EDIT: the binomial distribution is based on the sumscore. It would be the proper distribution if you were using $y = \sum_{i=1}^n x_i$.

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Thining about it I asked a yes/no quesiton so this does actually answer it. Thanks. –  Bowler Jan 16 '12 at 15:28