The following problem arises in the context of private equity, which typically report "smoothed" returns (think of it as a moving average). As you can imagine, "smoothed" returns would have a much lower volatility compared to the volatility of "unsmoothed" returns. For risk calculation we are interested in volatility of "unsmoothed" returns.

Mathematically, suppose I observe a process $\bar{r}_t$ which is a moving average of process $r_t$, i.e., $\bar{r}_t = \sum_{k=0}^p w_k r_{t-k}$. I also know that $r_t = \alpha + \beta r_{I, t} + \epsilon_t$, where $r_{I, t}$ are returns of a public index and $\epsilon_t = N(0, \sigma^2)$. I would like to estimate "unsmoothed" returns $r_t, t = 0, \ldots, T$ from the data: $\bar{r}_t, r_{I,t}, t=0, 1, \ldots, T$.

Can somebody suggest how should I go about this estimation? If there is a reference to a similar problem, that should be fine too. Thanks.


4 Answers 4


Did you try solving for $w_k$?

$$\bar{r}_t = \sum_{k=0}^p w_k r_{t-k}$$

$$\bar R = W R$$

Since you probably have $t>>k$, you can solve for $W$ using OLS $$\bar R = W R +\varepsilon$$


You can try applying Kalman filter. Here, your state evolution is $$r_t=\mu+\varepsilon_t$$. You introduce new vector $x_t=(r_t, r_{t-1}, \dots, r_{t-p+1})$ and $\mu_x=(\mu,\mu,\dots,\mu)$ re-write this as: $$x_t = \mu_x+x_{t-1}+e_t$$

Assuming the returns are independent and variance is constant AND that weights add up to 1, i.e. $\varepsilon\sim\mathcal{N}(0,\sigma)$ you can see that $\mu_{\bar r}=\mu_r$ and $\sigma_{\bar r}^2=\sigma_r^2\sum_{k=0}^{p-1}w_k^2$. Hence, $e_t$ is a multivariate normal with a diagonal covaraince matrix $diag(\Sigma)=\frac{1}{\sum_{k=0}^{p-1}w_k^2}(\sigma_{\bar r}^2,\sigma_{\bar r}^2,\dots,\sigma_{\bar r}^2)$.

Next, your measurement equation is $$\bar x_t=x_t \cdot(w_0,w_1,\dots,w_{p-1})'$$

This should be very easy to estimate using Kalman filter packages.

  • $\begingroup$ I neither know $w$ nor $r_t$, so it is not possible to do the OLS. That is why I have tried to relate $r_t$ to $r_{I,t}$. $\endgroup$
    – vdesai
    May 4, 2015 at 13:40
  • $\begingroup$ @vdesai, added Kalman filter suggestion $\endgroup$ May 4, 2015 at 18:02

Thanks @Aksakal for suggesting Kalman Filter. Here I provide more details. We will view it as a state-space model: $$ \begin{split} z_t &= A_t z_{t-1} + B_t u_t + \epsilon_t, \\ y_t &= C_t z_t + D_t u_t + \delta_t, \\ \epsilon_t &\sim \mathcal{N}(0, Q_t),\ \delta_t \sim \mathcal{N}(0, R_t), \end{split} $$ where $z_t$ is the latent variable, $y_t$ is the observation, $u_t$ is an optional input or control signal, $\epsilon_t$ is the system noise and $\delta_t$ is the observation noise.

Mapping our problem to state-space form, we get $$ z_t = \begin{bmatrix} r_t\\ r_{t-1} \\ \vdots \\ r_{t-p} \end{bmatrix}_{(p+1)\times 1}, A_t = \begin{bmatrix} 0, 0, \ldots, 0 \\ 1, 0, \ldots, 0 \\ \vdots \ldots \vdots \\ 0, \ldots, 1, 0 \end{bmatrix}_{(p+1)\times(p+1)}, B_t = \begin{bmatrix} \alpha,\ \beta \\ 0,\ 0 \\ \vdots\ \vdots \\ 0,\ 0 \end{bmatrix}, u_t = \begin{bmatrix} 1\\ r_{I, t} \end{bmatrix},\ Q_t = \begin{bmatrix} \sigma^2\quad &\mathbf 0^\top_{1\times p} \\ \mathbf{0}_{p \times 1}\ &\mathbf{0}_{p\times p} \end{bmatrix} $$ $$ y_t = \bar{r}_t,\quad \\ C_t = \begin{bmatrix} w_0\ w_1 \ldots w_p \end{bmatrix},\ D_t = \mathbf{0}_{1\times 2},\ R_t = \mathbf{0}_{1\times 1}. $$ We also assume $y_t \sim N(0, \sigma_0^2)$, where $\sigma_0^2$ is large enough so that it will be a diffuse prior. Our model has the parameter $\theta = (B_t, C_t, \sigma^2, \sigma_0^2)$. We are mainly interested in $p(z_t|y_{0\colon t}, u_{0\colon t}, \theta)$.


If you assume first order correlation and stationnary assumptions and no autocorrelation between true returns and estimated returns, the answer is the following

Denote by $ R^e $ the estimated return, $R^t $ the true return and $ \rho $ the autocorrelation coefficient

By assumptions, you have that

  • $ R^e(t)= \rho R^e(t-1) + (1-\rho ) R^t(t) $
  • $ Cov( R^t(t), R^e(t-1) ) = 0 $

Then the estimated volatility is

$Var(R^e(t))= Var\{\rho R^e(t-1)+(1-\rho)R^t(t)\} = \rho^2Var(R^e(t-1))+(1-\rho )^2Var(R^t(t))$ or

$Var(R^e)= \rho^2 Var( R^e) + (1-\rho )^2 Var(R^t) $


$Var(R^e)= \frac{1-\rho}{1+\rho} Var(R^t) $

this is the answer you may be looking for


The Blundell Ward filter is a fairly commonly used method for removing first order autocorrelation see;



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