RANDOM SIGNAL ANALYSIS X1 Xn are iid each with fxx and Fxx

RANDOM SIGNAL ANALYSIS:

X_1, ... X_n are iid, each with fx(x) and Fx(x). Z is the maximum of these N RVs. What is P[Z lessthanorequalto z]?

Solution

The log-likelihood function is l() = Xn i=1 \" log 2 log |Xi | # Let the derivative with respect to be zero: l 0 () = Xn i=1 \" 1 + |Xi | 2 # = n + Pn i=1 |Xi | 2 = 0 and this gives us the MLE for as ˆ = Pn i=1 |Xi | n Again this is different from the method of moment estimation which is ˆ = sPn i=1 X2 i 2n Example 3: Use the method of moment to estimate the parameters µ and for the normal density f(x|µ, 2 ) = 1 2 exp ( (x µ) 2 2 2 ) , 4 based on a random sample X1, · · · , Xn. Solution: In this example, we have two unknown parameters, µ and , therefore the parameter = (µ, ) is a vector. We first write out the log likelihood function as l(µ, ) = Xn i=1 · log 1 2 log 2 1 2 2 (Xi µ) 2 ¸ = n log n 2 log 2 1 2 2 Xn i=1 (Xi µ) 2 Setting the partial derivative to be 0, we have l(µ, ) µ = 1 2 Xn i=1 (Xi µ) = 0 l(µ, ) = n + 3 Xn i=1 (Xi µ) 2 = 0 Solving these equations will give us the MLE for µ and : µˆ = X and ˆ = vuut 1 n Xn i=1 (Xi X) 2 This time the MLE is the same as the result of method of moment. From these examples, we can see that the maximum likelihood result may or may not be the same as the result of method of moment. Example 4: The Pareto distribution has been used in economics as a model for a density function with a slowly decaying tail: f(x|x0, ) = x 0x 1 , x x0, > 1 Assume that x0 > 0 is given and that X1, X2, · · · , Xn is an i.i.d. sample. Find the MLE of . Solution: The log-likelihood function is l() = Xn i=1 log f(Xi |) = Xn i=1 (log + log x0 ( + 1) log Xi) = n log + n log x0 ( + 1)Xn i=1 log Xi Let the derivative with respect to be zero: dl() d = n + n log x0 Xn i=1 log Xi = 0 5 Solving the equation yields the MLE of : ˆMLE = 1 log X log x0