The discussion focuses on the natural logarithm of the likelihood function L(θ) defined as L(θ) = ∏(θ/(2√xi)*e^(-θ√xi)), for i=1 to n. Participants clarify that taking the natural log of a product results in the sum of the logs, leading to the expression lnL(θ) = nlnθ + Ʃln(1/(2√xi)*e^(-θ√xi)). The simplification process involves applying properties of logarithms, specifically ln(xa) = a ln(x) and ln(a/b) = ln(a) - ln(b), to further break down the components of the likelihood function.
PREREQUISITESStatisticians, data analysts, and students studying statistical inference who are interested in understanding likelihood functions and their applications in data analysis.
You can apply that again to the product [1/(2√xi)]*[e^(-θ√xi)]. You can then also use ln(xa) = a ln(x) for cases like a = -1, a = 1/2. What can you do with ln(ex)?Phox said:= nlnθ + Ʃln(1/(2√xi)*e^(-θ√xi))
Ok so from what I understand the ln of a product is the sum.
Phox said:Homework Statement
L(θ) = ∏(θ/(2√xi)*e^(-θ√xi)),i=1, n
Homework Equations
The Attempt at a Solution
-> θ2∏(1/(2√xi)*e^(-θ√xi))
taking natural log of both sides
lnL(θ) = nlnθ + ln∏(1/(2√xi)*e^(-θ√xi))
= nlnθ + Ʃln(1/(2√xi)*e^(-θ√xi))
Ok so from what I understand the ln of a product is the sum. But I'm not sure how exactly to simplify from here