Set
$$
a=e^{-(t_2-t_1)/\tau},
\qquad
b=e^{-(t_3-t_2)/\tau}.
$$
The <Markov factorization> and the conditional normal laws from part b give the fully univariate product
$$
p(y\mid t,\mu,\tau)
=\phi(y_1;\mu,1)
\phi\!\left(y_2;\mu+a(y_1-\mu),1-a^2\right)
\phi\!\left(y_3;\mu+b(y_2-\mu),1-b^2\right),
$$
where $\phi(\mathord\cdot;m,v)$ denotes the $N(m,v)$ density. This is a <weighted least squares> problem in $\mu$. Differentiating its log-likelihood gives
$$
\widehat\mu=
\frac{
y_1+\dfrac{y_2-ay_1}{1+a}+\dfrac{y_3-by_2}{1+b}
}{
1+\dfrac{1-a}{1+a}+\dfrac{1-b}{1+b}
}.
$$
Every innovation in the numerator has expectation equal to its coefficient in the denominator times $\mu$. Therefore $\mathbb E\widehat\mu=\mu$, so this maximum likelihood estimator is unbiased.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 219 / 2 / c / Solution

{c}

For an ordered Markov sequence, the joint density factors into an initial density and one-step transition densities:
$$
p(x_1,\ldots,x_n)=p(x_1)\prod_{i=2}^np(x_i\mid x_{i-1}).
$$


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