Let θ be an unknown constant. Let W1,…,Wn be independent exponential random variables each with parameter 1. Let Xi=θ+Wi.
What is the maximum likelihood estimate of θ based on a single observation X1=x1? Enter your answer in terms of x1 (enter as x_1) using standard notation .
θ^ML(x1)=
- unanswered
What is the maximum likelihood estimate of θ based on a sequence of observations (X1,…,Xn)=(x1,…,xn)?
θ^ML(x1,…,xn)=
- unanswered
(x1x2⋯xn)1/n
x1+⋯+xnn
11x1+⋯1xn
minixi
maxixi
None of the above
You have been asked to construct a confidence interval of the particular form [Θ^−c,Θ^], where Θ^=mini{Xi} and c is a constant that we need to choose. For n=10, how should the constant c be chosen so that we have a 95% confidence interval? (Give the smallest possible value of c.) Your answer should be accurate to 3 decimal places.
x_1
minXi
i
c=0.29957
To find the maximum likelihood estimate (MLE) of θ based on a single observation X1=x1, we need to find the value of θ that maximizes the likelihood function. The likelihood function L(θ|x1) is the probability density function of X1=x1, given θ.
Since X1 is the sum of θ and a single exponential random variable, we have:
X1 = θ + W1
where W1 is an exponential random variable with parameter 1.
The probability density function of W1 is given by:
f(W1) = e^(-W1), for W1 >= 0
Now, we can write the likelihood function as:
L(θ|x1) = f(x1-θ) = e^(-(x1-θ)), for x1 >= θ
To find the MLE, we maximize the likelihood function with respect to θ by taking the derivative and setting it to zero. However, in this case, we can easily see that the likelihood function is maximized when θ is equal to x1.
Therefore, the maximum likelihood estimate of θ based on a single observation X1=x1 is:
θ^ML(x1) = x1
For the second question, finding the MLE of θ based on a sequence of observations (X1,…,Xn)=(x1,…,xn), we can follow a similar approach. Each Xi is the sum of θ and an exponential random variable. The likelihood function in this case is given by:
L(θ|x1,…,xn) = f(x1-θ)f(x2-θ)...f(xn-θ)
= e^(-(x1-θ))e^(-(x2-θ))...e^(-(xn-θ))
= e^(-nx1+nθ)e^(-nx2+nθ)...e^(-nxn+nθ)
= e^(-n(x1+x2+...+xn) + n^2θ)
To maximize the likelihood function, we take the derivative with respect to θ and set it to zero:
d/dθ (L(θ|x1,…,xn)) = -n^2e^(-n(x1+x2+...+xn) + n^2θ) + n^2θe^(-n(x1+x2+...+xn) + n^2θ) = 0
Simplifying, we get:
- n^2 + n^2θ = 0
θ = 1/n(x1+x2+...+xn)
Therefore, the maximum likelihood estimate of θ based on a sequence of observations (X1,…,Xn)=(x1,…,xn) is:
θ^ML(x1,…,xn) = (x1+x2+...+xn)/n
Now, for the third question, we need to construct a confidence interval of the form [Θ^−c,Θ^] with a 95% confidence level, where Θ^ = min{Xi}.
In order to find the value of c, we need to consider the distribution of Θ^, which is the minimum of n exponential random variables. The distribution of Θ^ is also an exponential distribution with parameter n.
To construct a 95% confidence interval, we need to find the value of c such that P(Θ^−c ≤ Θ^) = 0.95.
Using the exponential distribution, we know that P(Θ^ > c) = e^(-nc).
Therefore, we have:
P(Θ^−c ≤ Θ^) = 1 - P(Θ^ > c)
= 1 - e^(-nc)
To achieve a 95% confidence level, we set this probability equal to 0.95:
1 - e^(-nc) = 0.95
Solving for c, we get:
e^(-nc) = 0.05
Taking the natural logarithm of both sides:
-nc = ln(0.05)
Solving for c, we have:
c = -ln(0.05)/n
For n=10, substituting into the equation, we find:
c = -ln(0.05)/10
Using a calculator, we can find the numerical value of c accurate to 3 decimal places:
c = 1.609
Therefore, the constant c should be chosen as 1.609 in order to have a 95% confidence interval for n=10.