I am learning about recursive least squares estimation using a forgetting factor $\lambda$ as a tool for treating time variations of model parameters and have become stuck on the following problem.
Question
Find an expression for $V\big[\hat{b}\big]$ given $$y_t = bu_t + e_t, \quad t=1,...,N$$
Where $e_t$ is white Gaussian noise with variance $\sigma^2_e$ and $u_t$ is a deterministic signal such that
$$\lim_{N\to\infty} \frac{1}{N} \sum_{t=1}^{N} u^2_t$$
is finite. The unknown parameter b is estimated as $$\hat{b}= \operatorname*{argmin}_b \sum_{t=1}^{N} \lambda^{N-t}(y_t-bu_t)^2,$$ where $0<\lambda \leq 1$.
My attempt at a solution
I can be seen that the argument that minimises the above equation is $\hat{b}= \frac{y_t}{u_t}$. However when I try to calculate the variance I get
$$V\big[\hat{b}\big]=V\big[\frac{y_t}{u_t}\big].$$ But as $u_t$ is a deterministic signal and I am under the impression that the variance of a deterministic signal is zero this would give me a zero in the denominator?
Any help greatly appreciated.
Edit
After user617446 comment I went back and recalculated $\hat{b}$ as follows-
$$\frac{\partial}{\partial b}\bigg[ \sum_{t=1}^{N} \lambda^{N-t}(y_t-bu_t)^2 \bigg] = 2b\sum_{t=1}^{N}\lambda^{N-t}u_t^2-2\sum_{t=1}^{N}\lambda^{N-t}y_tu_t$$
setting this equal to zero then solving gave
$$\hat{b}=\frac{\sum_{t=1}^{N}\lambda^{N-t}y_tu_t}{\sum_{t=1}^{N}\lambda^{N-t}u_t^2}.$$
I believe this to be correct but I am now stuck once again on how to calculate the variance? Grateful for any and all help.
Firstly, the least square estmation can be found via differentiation of sum by the parameter $\hat b,$ so the expression $$\hat b =\dfrac{\sum\limits_{t=1}^N\lambda^{N-t}y_t u_t}{\sum\limits_{t=1}^N\lambda^{N-t}u_t^2}$$ is correct.
Parameter $\hat b$ should be considered as random variable whose value depends on the specific white noise sample, $$\hat b =\dfrac{\sum\limits_{t=1^N}\lambda^{N-t}(e_t+u_t b) u_t}{\sum\limits_{t=1}^N\lambda^{N-t}u_t^2} =b + \dfrac{\sum\limits_{t=1}^N\lambda^{N-t}e_t u_t}{\sum\limits_{t=1}^N\lambda^{N-t}u_t^2}.$$ There are not reasons why the sums ratio can deviate the average mean of the random variable $\hat b,$ so $$M(\hat b) = b.$$ Then the variance is $$V(\hat b) = M((\hat b-b)^2) = M\left(\left(\dfrac{\sum\limits_{t=1}^N\lambda^{N-t}e_t u_t}{\sum\limits_{t=1}^N\lambda^{N-t}u_t^2}\right)^2\right)\\[4pt] = \dfrac{M\left(\sum\limits_{t=1}^N \lambda^{2(N-t)}u_t^2e_t^2\right) +M\left(\sum\limits_{1\leq t_1 < t_2\leq N} \lambda^{2N-t_1-t_2}u_{t_1}u_{t_2}e_{t_1} e_{t_2}\right)}{\left(\sum\limits_{t=0}^N\lambda^{N-t}u_t^2\right)^2} = \color{brown}{\mathbf{\dfrac{\sum\limits_{t=1}^N \lambda^{2(N-t)}u_t^2}{\left(\sum\limits_{t=1}^N\lambda^{N-t}u_t^2\right)^2}\cdot\sigma_e^2}}.$$