Let's consider independent random vectors $\hat{\boldsymbol\theta}_i$, $i = 1, \dots, m$, which are all unbiased for $\boldsymbol\theta$ and that $$\mathbb{E}\left[\left(\hat{\boldsymbol\theta}_i - \boldsymbol\theta\right)^{T}\left(\hat{\boldsymbol\theta}_i - \boldsymbol\theta\right)\right] = \sigma^2\text{.}$$ Let $\mathbf{1}_{n \times p}$ be the $n \times p$ matrix of all ones.

Consider the problem of finding $$\mathbb{E}\left[\left(\hat{\boldsymbol\theta} - \boldsymbol\theta\right)^{T}\left(\hat{\boldsymbol\theta} - \boldsymbol\theta\right)\right]$$ where $$\hat{\boldsymbol\theta} = \dfrac{1}{m}\sum_{i=1}^{m}\hat{\boldsymbol\theta}_i\text{.}$$

My attempt is to notice the fact that $$\hat{\boldsymbol\theta} = \dfrac{1}{m}\underbrace{\begin{bmatrix} \hat{\boldsymbol\theta}_1 & \hat{\boldsymbol\theta}_2 & \cdots & \hat{\boldsymbol\theta}_m \end{bmatrix}}_{\mathbf{S}}\mathbf{1}_{m \times 1}$$ and thus $$\text{Var}(\hat{\boldsymbol\theta}) = \dfrac{1}{m^2}\text{Var}(\mathbf{S}\mathbf{1}_{m \times 1})\text{.}$$ How does one find the variance of a random matrix times a constant vector? You may assume that I am familiar with finding variances of linear transformations of a random vector: i.e., if $\mathbf{x}$ is a random vector, $\mathbf{b}$ a vector of constants, and $\mathbf{A}$ a matrix of constants, assuming all are comformable, $$\mathbb{E}[\mathbf{A}\mathbf{x}+\mathbf{b}] = \mathbf{A}\mathbb{E}[\mathbf{x}]+\mathbf{b}$$ $$\mathrm{Var}\left(\mathbf{A}\mathbf{x}+\mathbf{b}\right)=\mathbf{A}\mathrm{Var}(\mathbf{x})\mathbf{A}^{\prime}$$

  • $\begingroup$ It isn't any different than finding any other variance: the matrix has $m^2$ components and therefore its variance (aka "variance-covariance matrix") is a quadratic form on $\mathbb{R}^{m^2}$. It's unclear, though, how your question is related to the situation described at the outset, which concerns finding the variance of the random vector (not matrix!) $\hat\theta$. $\endgroup$
    – whuber
    Commented Dec 22, 2017 at 16:33
  • $\begingroup$ @whuber I thought that writing $\hat{\boldsymbol\theta}$ as a product of a random matrix $\mathbf{S}$ (above) and a constant vector ($\mathbf{1}_{m \times 1}$) might make this computation cleaner. If this isn't the case, I'll just stick to traditional methods. $\endgroup$ Commented Dec 22, 2017 at 16:41
  • $\begingroup$ It very well could simplify some aspects of the computation (I haven't checked)--but you have noted the cost: there are a lot more components to keep track of. $\endgroup$
    – whuber
    Commented Dec 22, 2017 at 16:52

2 Answers 2


Using the matrix computation only (although essentially, this solution is not that different from @DeltaIV's direct calculation), let me first slightly modify your definition of $S$ to its centralized version $\begin{bmatrix}\hat{\theta}_1 - \theta & \cdots & \hat{\theta}_m - \theta\end{bmatrix}$. We can go as follows \begin{align} & E[(\hat{\theta} - \theta)^T(\hat{\theta} - \theta)] \\ = & \frac{1}{m^2}E[1^TS^TS1] \\ = & \frac{1}{m^2}1^T E[S^TS] 1 \tag{1} \\ = & \frac{1}{m^2}1^T \begin{bmatrix} E[(\hat{\theta}_1 - \theta)^T(\hat{\theta}_1 - \theta)] & \cdots & E[(\hat{\theta}_1 - \theta)^T(\hat{\theta}_m - \theta)] \\ \vdots & \ddots & \vdots \\ E[(\hat{\theta}_m - \theta)^T(\hat{\theta}_1 - \theta)] & \cdots & E[(\hat{\theta}_m - \theta)^T(\hat{\theta}_m - \theta)] \end{bmatrix} 1 \\ = & \frac{1}{m^2}1^T\text{diag}(\sigma^2, \ldots, \sigma^2)1 \tag{2} \\ = & \frac{1}{m}\sigma^2. \end{align}

In $(1)$, we used the fact that for any conformable non-random matrices $A$, $B$ and random matrix $X$, $E[AXB] = AE[X]B$.

In $(2)$, we applied the independence assumption.


Why involving random matrices? This seems much simpler:

$$\mathbb{E}\left[\left(\hat{\boldsymbol\theta} - \boldsymbol\theta\right)^{T}\left(\hat{\boldsymbol\theta} - \boldsymbol\theta\right)\right] = \frac{1}{m^2}\mathbb{E}\left[\sum_{i=1}^{m}\left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)^T\sum_{j=1}^{m}\left(\hat{\boldsymbol\theta}_j-\boldsymbol\theta\right)\right]= \frac{1}{m^2}\mathbb{E}\left[\sum_{i=1}^{m}\left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)^T\left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)\right] + \frac{1}{m^2} \mathbb{E}\left[\sum_{i=1}^{m}\sum_{j\neq i}^{m} \left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)^T\left(\hat{\boldsymbol\theta}_j-\boldsymbol\theta\right)\right]$$


$$\frac{1}{m^2}\mathbb{E}\left[\sum_{i=1}^{m}\left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)^T\left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)\right] = \frac{1}{m^2}\sum_{i=1}^{m}\mathbb{E}\left[\left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)^T\left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)\right]=\frac{m\sigma^2}{m^2}=\frac{\sigma^2}{m}$$

While, for $j\neq i$,


where the penultimate equivalence follows from the independence of $\hat{\boldsymbol\theta}'_i$ and $\hat{\boldsymbol\theta}'_j$, and the last equivalence follows from the fact that they are both zero mean.


$$\mathbb{E}\left[\left(\hat{\boldsymbol\theta} - \boldsymbol\theta\right)^{T}\left(\hat{\boldsymbol\theta} - \boldsymbol\theta\right)\right] = \frac{\sigma^2}{m}$$


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