MMSE Estimation - Fusion of 2 Measurements
*STOP! If you only want a hint and not the complete solution please see Stanley P.'s or Peter K.'s answers. *

Since you do not specify if there is model for the temperature evolving over time $n$, I will derive an estimator which is a combination of $Y_1$ and $Y_2$ for each fixed $n$. Let $\alpha \in (0,1)$ and suppose the estimate of $X$ can be written as $\hat X = \alpha Y_1 + (1-\alpha)Y_2$. The MSE is given by
\begin{eqnarray}
E[(\hat X-X)^2] &amp;=&amp; E[(\alpha Y_1 + (1-\alpha)Y_2 - X)^2]\\
&amp;=&amp; E[(\alpha X + (1-\alpha)X + \alpha V + (1-\alpha)W - X)^2]\\
&amp;=&amp; E[(\alpha V + (1-\alpha)W)^2]\\
&amp;=&amp; \alpha^2 \sigma_v^2 + (1-\alpha)^2\sigma_w^2.
\end{eqnarray}
To minimize this MSE, take the derivative with respect to $\alpha$ and set it to zero. After some algebra you get $\alpha = \frac{\sigma_w^2}{\sigma_v^2+\sigma_w^2}$ and so
$$
\hat X = \frac{\sigma_w^2}{\sigma_v^2+\sigma_w^2} Y_1 + \frac{\sigma_v^2}{\sigma_v^2+\sigma_w^2 }Y_2.
$$

Note that if the variances are unequal (say $\sigma_w^2 &gt; \sigma_v^2$), then the noisier observation ($Y_2$) is given less weight than the less noisy observation ($Y_1)$. If the noise variances are equal, the two observations are weighted equally and we get just the arithmetic mean of the two observations.

If you do have a model for temperature evolution over time you should use a Kalman filter which is guaranteed to produce the MMSE estimate.