I didn’t understand this question. How change in beta affects the location of the red line?

Increasing beta will give higher weight to data at previous timestep. This means that as we increase beta, the curve will shift towards the right since you’re paying more attention to what was a timestep earlier.

So, the 1st option shouldn’t be selected.

Another way to think is, if beta = 0, then, v(t) = theta(t) i.e. the curve remains wrt time.

If beta = 1, v(t) = v(t-1) i.e. shifted right.

@balaji.ambresh has it right with *Increasing beta will give higher weight to data at previous timestep*. But I found it difficult to visualize how that translates into shape of the curve. Here’s a visualization I made using random noise data (not the actual London data referenced in the quiz)

The dots are the raw data. The red line is \beta = 0.4. The green line is \beta = 0.9. And the blue line is \beta = 1.0.

\beta weights the contribution of the previous time period. At the lower limit of \beta = 0, there is no contribution of the previous period, and the weighted line tracks the raw data exactly. At the upper limit of \beta = 1, each value is exactly equal to the previous value, in other words a flat line. As \beta increases the curve of v_t becomes flatter, which means it reaches a given value at a later period (which is what is meant by moving to the right). As \beta decreases the curve of v_t tracks \theta_t more closely and reaches a given value sooner (moving to the left). Hope this helps.