MLE

• For any single attempt, what is the probability of picking the right key? (Your answer isn’t a number, but an expression.)
• If \( X \) is the random variable that gives the number of attempts to grab the right key, what is \( P(X=8) \)?
• What is the probability of obtaining the values of \( X \) listed above: 8, 12, 7, 6, 12? (Give your best answer, this really requires a course in probability, but I bet you get it.)

• \( p=\frac{1}{N} \)
• \( P(X=8)=(1-p)^7\cdot p \).
• That is seven failures, followed by one success.
• \[ P(X=8)\cdot P(X=12) \cdot P(X=7)\cdot P(X=6) \cdot P(X=12) \]
• That is, we multiply the probabilities: this works as long as each trial is independent of the others.

• \( p \) is the parameter in this distribution
• We don’t know the value of \( p \) because we don’t know \( N \).
• We’d like to find the value of \( p \) that maximizes the probability of obtaining the observed result.
• How? Calculus.

• Define the function: \[ L(\theta | x_1, x_2, \dots, x_n)=P(X_1=x_1) \cdot P(X_2=x_2) \cdots P(X_n=x_n) \]
• Think of this as a function of the parameter \( \theta \) (\( p \) for this particular example) where the \( x_i \) s are the observed values of the random variable.
• Write down \( L(\theta|x_1, \dots x_n) \) for this example. Your function should be a function of \( p \) and the observed values should appear in your function.

\[ \begin{align} &L(p | 8, 12, 7, 6, 12)\\ &=(1-p)^7 \cdot p \cdot (1-p)^{11} \cdot p \cdot (1-p)^6 \cdot p\\ &\cdot (1-p)^5 \cdot p \cdot (1-p)^{11} \cdot p \end{align} \]

• We’d like to maximize this function!
• How does that go?

• So, taking the derivative of \( L(p) \) doesn’t look pleasant - primarily because this is a product and we’ll either have to expand \( L(p) \) or apply the product rule…. a lot.
• However, \( \ln \) turns products into sums: \( \ln(A\cdot B) = \ln(A) +\ln(B) \)
• …. and the same value that maximizes \( \ln(L(p)) \) maximizes \( L(p) \)

• Take \( \ln \) of: \[ \begin{align} &L(p)\\ &=(1-p)^7 \cdot p \cdot (1-p)^{11} \cdot p \cdot (1-p)^6 \cdot p\\ &\cdot (1-p)^5 \cdot p \cdot (1-p)^{11} \cdot p \end{align} \]
• If you’re comfortable with the following notation, you can write it with summation notation: \[ L(p)=\prod_{i=1}^n (1-p)^{x_i-1}\cdot p \]

\[ \begin{align} \ln(L(p))&=\sum_{i=1}^n \ln(1-p)^{x_i-1} + \sum_{i=1}^n \ln(p)\\ &=\sum_{i=1}^n (x_i-1) \cdot \ln(1-p) + \sum_{i=1}^n \ln(p)\\ &= \ln(1-p) \sum_{i=1}^n (x_i-1) + \ln(p) \cdot \sum_{i=1}^n 1\\ &= \ln(1-p) \sum_{i=1}^n (x_i-1) + \ln(p)\cdot n \end{align} \]

• What is \( n \)?
• What is \( \sum_{i=1}^n (x_i-1) \)?

• \( n=5 \)
• \( \sum_{i=1}^n (x_i-1)=7+11+6+5+11=40 \)
• So we have \[ \ln(L(p))= 40\ln(1-p) + 5\ln(p) \]
• Now maximize \( \ln(L(p)) \).

\[ \frac{d}{dp} \ln(L(p))=-40\cdot \frac{1}{1-p}+\frac{5}{p} \] Set equal to zero: \[ \begin{align} 0&=-40\cdot \frac{1}{1-p}+\frac{5}{p}\\ \frac{40}{1-p}&=\frac{5}{p}\\ \frac{1-p}{40}&=\frac{p}{5}\\ 1-p&=8p\\ 1&=9p\\ p&=\frac{1}{9} \dots \text{ so } N=9\\ \end{align} \]

• This is called the maximum likelihood estimator process.
• We maximized \[ L(\theta | x_1, x_2, \dots, x_n)=P(X_1=x_1) \cdot P(X_2=x_2) \cdots P(X_n=x_n) \]
• We replaced \( P(X_i=x_i) \) with \( (1-p)^{x_i-1}\cdot p \) which is the pmf for the geometric distribution.
• So, \[ L(\theta | x_1, x_2, \dots, x_n)=f(x_1, \theta) \cdot f(x_2, \theta) \cdots f(x_n, \theta) \] where \( f \) is either the pmf or the pdf depending on whether the random variable is discrete or continuous.
• We find the value of the parameter, \( \theta \), that maximizes \( L(\theta) \).
• Typically by instead maximizing \( \ln(L(\theta)) \) since that tends to be easier.