Let’s start with the very simple case where we have one series $y$ with 10 independent observations: 5, 0, 1, 1, 0, 3, 2, 3, 4, 1.
We see from this that the sample mean is what maximizes the likelihood function. Using the given sample, find a maximum likelihood estimate of \(\mu\) as well. 7), then we cannot use differentiation and we need to find the maximizing value in another way. e. Since the maximum likelihood estimator
maximizes the log-likelihood, it satisfies the first order
conditionFurthermore,
by the Mean Value Theorem, we
havewhere,
for each
,
the intermediate points
satisfyand
the
notationindicates
that each row of the Hessian is evaluated at a different point (row
is evaluated at the point
).
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Continuous variables. The above example gives us the idea behind the maximum likelihood estimation. We now read here how the former can
be weakened and how the latter can be made more specific. We now would like to talk about a systematic way of parameter estimation. Conveniently, most common probability distributions – in particular the exponential family – are logarithmically concave.
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Also, note that the increase in \(\log \mathcal{L}(\boldsymbol{\beta}_{(k)})\)
becomes smaller with each iteration. (So, do you see from where the name “maximum likelihood” comes?) So, that is, in a nutshell, the idea behind the method of maximum likelihood estimation. The latent variables follow a normal distribution such that:$$y^* = x\theta + \epsilon$$
$$\epsilon \sim N(0,1)$$where$$ y_i = \begin{cases} 0 \text{ if } y_i^* \le 0\\ 1 \text{ if } y_i^* \gt 0\\ \end{cases} $$The probability density $$P(y_i = 1|X_i) = P(y_i^* \gt 0|X_i) = P(x\theta + \epsilon\gt 0|X_i) = $$
$$P(\epsilon \gt -x\theta|X_i) = 1 – \Phi(-x\theta) = \Phi(x\theta)$$where $\Phi$ represents the normal cumulative distribution function. }
Theoretically, the most natural approach to this constrained optimization problem is the method of substitution, that is “filling out” the restrictions
h
1
,
h
2
,
,
h
r
{\displaystyle \;h_{1},h_{2},\ldots ,h_{r}\;}
to a set
h
1
,
h
2
,
,
h
r
,
h
r
+
1
,
,
h
k
{\displaystyle \;h_{1},h_{2},\ldots ,h_{r},h_{r+1},\ldots ,h_{k}\;}
in such a way that
h
=
[
h
1
,
h
2
,
,
h
k
]
{\displaystyle \;h^{\ast }=\left[h_{1},h_{2},\ldots ,h_{k}\right]\;}
is a one-to-one function from
R
k
{\displaystyle \mathbb {R} ^{k}}
to itself, and pop over here the likelihood function by setting
i
=
h
i
(
1
,
2
,
,
k
)
.
Dear This Should you can try here Density Function
.