Question

It is the first day of spring, and a gloomy 100-day winter has finally ended. Over those 100 days, 36 of them were clear, 24 of them were cloudy (but not rainy), and the remaining 40 were rainy. Denote the probability that a given winter day next year will be clear, cloudy, and rainy, respectively, and assume that next winter’s weather will follow the same pattern as this year’s. Please derive expressions for the maximum-likelihood estimates of Pclear, Pcloudy and Prainy, in terms of this winter’s weather data.

Answer 1

P(clear) = no. of clear days / total days

= 36/100 = 0.36

P(cloudy) = no. of cloudy days / total days

= 24/100 = 0.24

P(rainy) = no. of rainy days / total days

= 40/100 = 0.40

maximum-likelihood estimates :

µ = n*p

σ = [n*p*(1-p)]^0.5

x = no. of days out of 100

for clear :

x = clear days in winter out of 100

µ = 100*0.36 = 36

σ = [100*0.36*0.64]^0.5 = 4.8

P(x clear days) = (1 / (4.8*2π)) * exp(-(x-36)^2 / (2*(4.8)^2))

= (1 / (9.6π)) * exp(-(x-36)^2 / (46.08))

P(x clear days) = (1 / (9.6π)) * exp(-(x-36)^2 / (46.08))

for cloudy :

x = cloudy days in winter out of 100

µ = 100*0.24 = 24

σ = [100*0.24*0.76]^0.5 = 4.2708

P(x cloudy days) = (1 / (4.2708*2π)) * exp(-(x-24)^2 / (2*(4.2708)^2))

= (1 / (8.5416π)) * exp(-(x-24)^2 / (36.4795))

P(x cloudy days) = (1 / (8.5416π)) * exp(-(x-24)^2 / (36.4795))

for rainy :

x = rainy days in winter out of 100

µ = 100*0.40 = 40

σ = [100*0.40*0.60]^0.5 = 4.8990

P(x rainy days) = (1 / (4.8990*2π)) * exp(-(x-40)^2 / (2*(4.8990)^2))

= (1 / (9.798π)) * exp(-(x-40)^2 / (48.0004))

P(x rainy days) = (1 / (9.798π)) * exp(-(x-40)^2 / (48.0004))

(please UPVOTE)