MLE for the probability of a dry day

Question

Daily weather is recorded as sun, cloud, or rain, treated as independent draws with fixed probabilities summing to 1. Over 200 days the counts are 120 sun, 48 cloud, 32 rain. Define a "dry" day as any day that is not rain.

What is the maximum likelihood estimate of the probability of a dry day?

Accepted Answer

The multinomial MLE for each category is its observed frequency, \hat pj = nj / n. A dry day is sun or cloud, so its probability is p{	ext{dry}} = p{	ext{sun}} + p{	ext{cloud}} = 1 - p{	ext{rain}}, a function of the category probabilities. By the invariance property of maximum likelihood, plug in the estimates: \hat p{	ext{dry}} = \frac{120 + 48}{200} = \frac{168}{200} = 0.84. Equivalently 1 - \hat p{	ext{rain}} = 1 - 32/200 = 0.84. Takeaway. Any pooled or complementary probability is estimated by pooling the corresponding counts, a direct use of MLE invariance.

MLE for the probability of a dry day

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