As we learned, “exponentially weighted averages” is useful to see a trend from noisy data. As you see, given equations are;

v_0 = 0, \ \ \ v_t = \beta v_{t-1} + (1-\beta)\theta_t, \ \ \ \beta = 0.5

Let’s think about the case just using the above equation.

v_0 = 0 \\
v_1 = \beta v_0 + (1-\beta)\theta_1 = 0.5*0 + 0.5*10 = .... \\
v_2 = \beta v_1 + (1-\beta)\theta_2 = 0.5*v_1 + 0.5*10 = ....

Using “weights”, we calculate “temperature” for a trend analysis from yesterday and today’s temperature. I think you can get the answer from the above easily.

As we started with v_0 = 0, this trend curve started from a lower value since it is an exponentially weighted average of \theta_1 and 0. We need to think about a bias term.

Andrew proposed a following equation for Bias correction, \frac{v_t}{1-\beta^t} instead of vt.

So, the above equations are now;

v_0 = v_0^{corrected} = 0, \ \ \ v_t^{corrected} =\frac{ \beta v_{t-1}^{correctetd} + (1-\beta)\theta_t}{1-\beta^t}, \ \ \ \beta = 0.5

Then, you can calculate “temperature” just like we did without bias correction.

So, there is the answer.

But, I realized that confusions may come from a “prime sign” that you pointed out. This is used in different ways in math, like “transpose”, “derivative”, .

In this case, I suppose this is either typo or problem in interpreting LaTex text in a browser. This must be “comma” to separate v_2 and v_2^{corrected}.

Sorry about this misleading prime sign…