# State Space models: rewriting the Likelihood to estimate the covariance matrix

I have a State Space model

$$\begin{matrix} Y_t & = & FX_t +R_{t}^{1/2}\epsilon_t \\ X_{t+1} & = & GX_{t}+Du_t \end{matrix}$$

where $$\epsilon_t$$ and $$u_{t}$$ distributed as $$N(0,I)$$ so that $$R_{t}^{1/2} \epsilon_{t}$$ is distributed as $$N(0,R_{t})$$. Notice the time index at $$R_{t}$$, as $$R_{t}$$ is a time-varying covariance matrix (suppose pd for simplicity). My question is: does it make any difference to compute and maximize the likelihood of $$Y_{t}$$ with respect to $$R_{t}$$ or maximize the likelihood of $$R_{t}^{-1/2}Y_{t}$$ with respect to $$R_{t}$$? The difference is that, in the latter case, you can conveniently set your covariance at the unit matrix I and write the negative log likelihood for $$R_{t}^{-1/2}Y_{t}$$ (given the parameters set $$\Theta$$ and $$X_{t}$$) to be minimized as:

$$L(R_{t}^{-1/2}Y_{t} | \Theta , X_{t} )= \sum_{t=1}^{T} log(|I|)+(R_{t}^{-1/2}Y_{t}-R_{t}^{-1/2}GX_{t})^{T}I(R_{t}^{-1/2}Y_{t}-R_{t}^{-1/2}GX_{t})$$

$$L(Y_{t} | \Theta , X_{t} )= \sum_{t=1}^{T} log(|R_{t}|)+(Y_{t}-GX_{t})^{T}R_{t}^{-1}(Y_{t}-GX_{t})$$

and minimize with respect to $$R_{t}$$. It seems what they are doing at pag. 22 here to simplify a similar problem. But I am not getting what happens to the term $$log(|I|)$$ which is indeed not comparable to $$log(|R_{t}|)$$. Are they 100% equivalent? What am I missing?

Thanks

I will try to give my interpretation here. I take your question as: "given the optimal $$\Sigma_{t}$$, does this value minimize the two log likelihoods simultaneously?" I will try to show so in an asymtptotically large sample of size $$T \rightarrow +inf$$.

Let's drop the conditional mean $$GX_{t}$$ to simplify the notation. So restate the model as

$$Y_{t}=R_{t}^{1/2} \epsilon_{t}$$

and assume that the true value of the unknown parameter $$R_{t}$$ to be estimated is $$A_{t}$$ so that the true process is $$Y_{t}=A_{t}^{1/2} \epsilon_{t}$$, which implies that $$A_{t}^{-1/2}Y_{t}=\epsilon_{t}$$ and, therefore, $$E_{t-1}(A_{t}^{-1/2}Y_{t})=E(A_{t}^{-1/2}Y_{t})=0$$. For an asympthotically large sample, the above mentioned properties should hold in the sample when its size approaches infinite and the unknown optimal value of the parameter to be estimated $$R_{t}$$ approaches $$A_{t}$$.

Indeed, we know from the consistency of the MLE that, IF the model is correctly specified, then the MLE estimate converges in probability to the true value of the unknown parameter $$A_{t}$$. Clearly, if that is the true value of the paramter, it must be the true value for both the processes $$Y_{t}$$ and $$A_{t}^{-1/2}Y_{t}$$ maximizing their repsective likelihood functions.

Now let's pause for a minute and consider that, if the same parameter $$\theta$$ simultaneously minimize two functions, call them b and c, then the following happens: when the parameter is at optimum and b is at the corresponding minimum value with respect to that paramter $$min(b)_{\theta}$$, then c is simultaneously minimized with respect to the same parameter iff the difference $$min(b)_{\theta}-min(c)_{\theta}$$ is minimum. And, analogously, iff $$min(c)_{\theta}-min(b)_{\theta}$$ is maximum. This will be useful later. Let's go on.

Given the function

$$L(Y_{t} | \Theta , X_{t} )= \sum_{t=1}^{T} log(|R_{t}|)+Y_{t}^{T}R_{t}^{-1}Y_{t}$$

the last term can be re-written as

$$Y_{t}^{T}R_{t}^{-1}Y_{t}= Tr(R_{t}^{-1}Y_{t}Y_{t}^{T}))= Tr(R_{t}^{-1} (E_{t-1}(Y_{t}Y_{t}^{T}) + WN_{t}))= Tr(R_{t}^{-1}E_{t-1}(Y_{t}Y_{t}^{T}))+Tr(WN_{t})$$ where $$WN_{t}$$ is a White Noise with unconditional mean of 0 when the sample size tends to +inf and the parameter values converge at their true value. Therefore, for an asympthotically large sample, where the optimal value of the unknown parameter converges to the true value, it is easy to show that $$\sum_{t} Tr(WN_{t})$$ goes to 0. Since we know that at the optimum this term must converge to 0, we can minimize the reminder of the function with respect to the unknown parameters. The same reasoning applies in case you work with the standardized variable $$R_{t}^{-1/2}Y_{t}$$. So we minimize

$$\sum_{t=1}^{T} log(|R_{t}|)+Tr(E_{t-1}( R_{t}^{-1}Y_{t}Y_{t})^{T})$$ and $$\sum_{t=1}^{T} log(|I|)+Tr(E_{t-1}( I^{-1}R_{t}^{-1/2}Y_{t}(R_{t}^{-1/2}Y_{t})^{T})$$

with respect to $$R_{t}$$ which must be pd. We call these functions -loglik1 and -loglik2. Due to the consistency of MLE, we know that asympthotically for $$T \rightarrow +inf$$ then the optimum will be $$R_{t}=A_{t}$$ for each t. Which implies that the first function value at the minimum is (by substitution)

$$\sum_{t=1}^{T} log(|A_{t}|)+Tr(E_{t-1}( A_{t}^{-1}Y_{t}Y_{t})^{T}) = \sum_{t=1}^{T} log(|A_{t}|)+Tr(A_{t}^{-1} E_{t-1}(Y_{t}Y_{t})^{T}) = \sum_{t=1}^{T} log(|A_{t}|)+Tr(I)$$

Before going one, observe that $$R_{t}=A_{t}$$ minimizes -loglik1 subject to $$det(R_{t})>0$$. Indeed, provided that the true process of $$Y_{t}$$ is $$Y_{t}=A^{1/2}_{t}\epsilon_{t}$$ so that $$E_{t-1}(Y_{t}Y_{t})^{T})=A_{t}$$,

$$\sum_{t=1}^{T} log(|R_{t}|)+Tr(E_{t-1}( R_{t}^{-1}Y_{t}Y_{t})^{T})= \sum_{t=1}^{T} log(|R_{t}|)+Tr( E_{t-1}(R_{t}^{-1}) E_{t-1}(Y_{t}Y_{t})^{T}) )= \sum_{t=1}^{T} log(|R_{t}|)+Tr( R_{t}^{-1}A_{t})$$

so taking the gradient with respect to $$R_{t}$$ and setting it equal to $$0$$ givesn the first order condition $$R_{t}=A_{t}$$ for each t, where $$det(A_{t})>0$$ by assumption (so $$R_{t}=A_{t}$$ minimizes the -loglik1 subject to $$det(R_{t})>0$$).

Provided that $$R_{t}=A_{t}$$ minimizes -loglik1 subject to $$det(R_{t})>0$$ due to the consistency of MLE and this is the minimum for -loglik1, then the parameter value $$R_{t}=A_{t}$$ will minimize also -loglik2 if the difference $$min(-loglik2)_{R_{t}}-min(-loglik1)_{R_{t}}$$ is maximum at $$R_{t}=A_{t}$$. Therefore, take the value of -loglik2 when $$R_{t}=A_{t}$$ (which is our expected minimum for $$T \rightarrow +inf$$ under the assumption that MLE is consistent). We derive it through the same passages adopted above for -loglik1:

$$\sum_{t=1}^{T} log(|I_{t}|)+Tr(I)$$

Notice that the difference $$min(-loglik2)_{R_{t}}-min(-loglik1)_{R_{t}}$$ is

$$\sum_{t=1}^{T} log(|I_{t}|)+Tr(I) - \sum_{t=1}^{T} log(|A_{t}|)+Tr(I) = - \sum_{t=1}^{T} log(|A_{t}|) = - log(\prod_{t=1}^{T}|A_{t}|) = - log(\prod_{t=1}^{T} \prod_{i=1}^{N}eig_{i}(A_{t}))$$

as the determinant can be re-expressed as the product of the N eigenvalues. Now, assuming that you are working with real $$NxN$$ square symmetyric pd matrixes whose eigenvalues are all real and positive (we are assuming those matrixes to be positive definite) and whose determinant is lower than 1 (*see at the end), then you have that, for $$T \rightarrow +inf$$, the term $$log( prod_{t=1}^{T}prod_{i=1}^{N}eig_{i}(A_{t}))$$ converges to $$-inf$$, which clearly implies that its additive inverse $$- log(\prod_{t=1}^{T} \prod_{i=1}^{N}eig_{i}(A_{t})) \rightarrow +inf$$ and is therefore maximized.

Therefore, if we assume that the true value of the parameter is the optimum for -loglik1, then, since this value is such that $$min(-loglik2)_{R_{t}}-min(-loglik1)_{R_{t}}$$ is max (diverges to $$+inf$$ as $$T \rightarrow +inf$$), we can conclude that the same value of the parameter $$R_{t}=A_{t}$$ minimizes -loglik2 as well.

(*) The covariance matrix can be represented as the product of 1) a diagonal matrix with standard deviations, 2) the corr matrix 3) again the diagonal matrix with standard deviations. Notice that its determinant is the product of the determinants of those 3 matrixes. Notice also that the determinant of the correlation matrix will be 1 iff such matrix is diagonal, less than 1 otherwise (see here), and notice that the determinant of the stdev matrixes will be less than 1 if the sum of elements of rows is less than 1 (as stated here), so if you have reasonably low values of standard deviations (like stdev < 100%), then you are in this scenario where the determinant of each conditional covariance matrix is less than 1 and $$log(det(\Sigma_{t}))$$ is negative as a consequence.