Question
Suppose that K = 3, and let E =\{ 1, 2, 3\}. Let X_1, \ldots , X_ n \stackrel{iid}{\sim } \mathbf{P}_{\mathbf{p}} for some unknown \mathbf{p} \in \Delta _3. Let f_{\mathbf{p}} denote the pmf of \mathbf{P}_{\mathbf{p}} and recall that the likelihood is defined to be
L_ n(X_1, \ldots , X_ n, \mathbf{p}) = \prod _{i = 1}^ n f_{\mathbf{p}}(X_ i).
Here we let the sample size be n = 12, and you observe the sample \mathbf{x} = x_1, \ldots , x_{12} given by
\mathbf{x} = 1, 3, 1, 2, 2, 2, 1, 1, 3, 1, 1, 2,.
The likelihood for this data set can be expressed as L_{12}( \mathbf{x}, \mathbf{p}) = p_1^ A p_2^ B p_3^ C.
Fill in the values of A, B, and C below.
A =
B =
C =
L_ n(X_1, \ldots , X_ n, \mathbf{p}) = \prod _{i = 1}^ n f_{\mathbf{p}}(X_ i).
Here we let the sample size be n = 12, and you observe the sample \mathbf{x} = x_1, \ldots , x_{12} given by
\mathbf{x} = 1, 3, 1, 2, 2, 2, 1, 1, 3, 1, 1, 2,.
The likelihood for this data set can be expressed as L_{12}( \mathbf{x}, \mathbf{p}) = p_1^ A p_2^ B p_3^ C.
Fill in the values of A, B, and C below.
A =
B =
C =
Answers
GPT 3.5
To calculate the likelihood, we need to count the occurrences of each value in the sample.
From the given sample, we have:
- The value 1 appears 6 times.
- The value 2 appears 4 times.
- The value 3 appears 2 times.
Therefore, we can calculate the likelihood as:
L_{12}( \mathbf{x}, \mathbf{p}) = p_1^6 p_2^4 p_3^2.
So, A = 6, B = 4, and C = 2.
From the given sample, we have:
- The value 1 appears 6 times.
- The value 2 appears 4 times.
- The value 3 appears 2 times.
Therefore, we can calculate the likelihood as:
L_{12}( \mathbf{x}, \mathbf{p}) = p_1^6 p_2^4 p_3^2.
So, A = 6, B = 4, and C = 2.
Related Questions
Let \widehat{\mathbf{p}} denote the MLE for a categorical statistical model ( \{ a_1, \ldots , a_ K...
Let X_1, \ldots , X_ n \stackrel{iid}{\sim }X\sim \mathbf{P} for some unknown distribution \mathbf{P...
As in the previous problem, we consider the matrix
H = I_ n - \frac{1}{n} \mathbf{1} \mathbf{1}^...
True or False:
\text {KL}(\mathbf{P}||\mathbf{Q})\geq 0 for all discrete distributions \mathbf{P}...