Answer to Question #148869 in Statistics and Probability for Parag Bhendare

Question

Q-3 The daily COVID 19 cases (in hundred) for Delhi for past 2 week is summarizing in the following table:
Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Cases 28 29 33 31 37 34 36 43 41 32 34 37 39 32

a. Using exponential smoothing method forecast the cases for 15 days, taking alpha as 0.3 and Initial forecast is the average of all data.
b. Using linear trend analysis, find the trend line for number of COVID 19 cases in Delhi and forecast for next 3 days. Also compute the Mean Square Error.

Answer 1

a). We remind that the formula for the forecast in exponential smoothing method has the form(https://towardsdatascience.com/simple-exponential-smoothing-749fc5631bed):

$f_t=\alpha d_{t-1}+(1-\alpha)f_{t-1}$

We set $\alpha=0.3$ and receive the following:

We will consider two cases:

1. We start from the first day. I.e., $d_1=28$, $f_1=\frac1{14}\sum_{i=1}^{14}d_i$ and we will use the formula for the rest.

We will get the following values for the first 15 days:

For the 15th day we received $f_{15}=35,4386$.

2.We can also set $d_1=32$ (value for the last 14th day) and $f_1=\frac1{14}\sum_{i=1}^{14}d_i=34.7143$

Then, $f_2=0.3\cdot32+0.7\cdot34,7143=33,9$. This is the forecast for the 15th day.

b). We remind that the aim is to find such line $l_t=at+b$ that $S=\sum_{t}[y_t-(at+b)]^2$ is minimized. After substitution of all values we receive: $S=17120-7512a-972b+1015a^2+210ab+14b^2$

We solve the system $\frac{\partial S}{\partial a}=2030a+210b-7512=0$ , $\frac{\partial S}{\partial b}=28b+210a-972=0$ . The latter is solved via: $a=\frac{222}{455}$ , $b=\frac{2826}{91}$ .

It is the global minimum.

We substitute the values $t=15,$ $t=16,$ $t=17$ in $at+b$ and receive:

$l_{15}\approx38.374$ ; $l_{16}\approx38.862$ ; $l_{17}\approx39.35$ . It is the approximation for the next 3 days.

The Mean Square Error is: $S=\frac{1}{14}\sum_{t=1}^{14}[y_t-(\frac{222}{455}t+\frac{2826}{91})]^2=\frac{44294}{3185}\approx13.91$

The latter was computed via the following formula in Maple:

S:=evalf(((28-(a*1+b))^2+(29-(a*2+b))^2+(33-(a*3+b))^2+(31-(a*4+b))^2+(37-(a*5+b))^2+(34-(a*6+b))^2+(36-(a*7+b))^2+(43-(a*8+b))^2+(41-(a*9+b))^2+(32-(a*10+b))^2+(34-(a*11+b))^2+(37-(a*12+b))^2+(39-(a*13+b))^2+(32-(a*14+b))^2)/14);