# Is there a case where Var[aX] = a*Var[X] where a is not equal to 0 and 1

Can someone explain how the equation in the below image is possible?

As far as I remember $Var[aX] = a^2.Var[X]$

You can access the image • You could have $a=1$. – Lord Shark the Unknown Dec 26 '17 at 5:42
• I will make the edit in the Question. Kindly take a look at the image. – Anirudh Murali Dec 26 '17 at 5:45
• This happens exactly when a=0, a=1 if Var(X)=0. – Kavi Rama Murthy Dec 26 '17 at 5:45
• Kindly take a look at the image. – Anirudh Murali Dec 26 '17 at 5:46
• I think the image may have a mistake. – BruceET Dec 26 '17 at 7:01

## 2 Answers

I assume you are asking why $\sigma_x^2 = 4.33 \cdot \sigma_w^2$ instead of $\sigma_x^2 = 4.33^2 \cdot \sigma_w^2$ when $x = 4.33w$. You are correct that $\text{Var}[aX] = a^2\text{Var}[X]$, but this equation does not apply here. The key difference is that we are not taking a single random week and and multiplying its forecast by $4.33$ - we are adding $4.33$ independent weeks together. To see the difference another way, suppose $X_1$ and $X_2$ are random variables that come from the same probability distribution. Then obviously we must have $\text{Var}[X_1] = \text{Var}[X_2]$. But note that $\text{Var}[2X_1] = 4 \text{Var}[X_1] = 4 \text{Var}[X_2]$ is different from $\text{Var}[X_1] + \text{Var}[X_2] = 2\text{Var}[X_1] = 2\text{Var}[X_2]$. To repeat, the main point here is that $\text{Var}[2X]$ is different from $\text{Var}[X] + \text{Var}[X]$. The second part (the summation) is what is used in the equation in the image you have provided. This is why your variance formula does not apply.

As a practical example to convince yourself, let $X$ be the random variable obtained by rolling a fair six-sided die. You should try calculating its variance by hand. Then try to calculate the variance of $2X$ (the random variable where you take twice the value obtained from rolling a fair six-sided die). This is $\text{Var}[2X]$. Now try calculating the variance of $Y$, where $Y$ is the random variable obtained as the sum of the rolls of two fair six-sided dice. You will find that this variance will be $2\text{Var}[X]$

• Thank you very much. I understand very well now. :) – Anirudh Murali Dec 26 '17 at 6:07

The text says $x = 4.33 w$, where $w$ is a weekly forecast, and $x$ is the average monthly forecast. This is correct but it obscures the fact that the monthly forecast would be in some sense the sum of independent and identically distributed weekly forecasts. Thus, what they are really trying to say is $$\operatorname{E}[X] = 4.33 \operatorname{E}[W],$$ and the variances are additive: $$\sigma_x^2 = 4.33 \sigma_w^2.$$

To give you an analogy, suppose that the daily forecast is a random variable $D$ with mean $\mu$ and variance $\sigma_D^2$. Then a weekly forecast would not simply be $W = 7D$. Instead, it should be $$W = D_1 + D_2 + \cdots + D_7,$$ where each $D_i \sim D$ for $i = 1, 2, \ldots$. This is because if I want to forecast the whole week, I need to add up the outcomes for every day in that week, rather than multiply the first day's outcome by $7$. But the expectation of the forecast does obey this simple multiplicative behavior: $$\operatorname{E}[W] = \operatorname{E}[D_1 + \cdots + D_7] = 7\operatorname{E}[D].$$ Now, if you recall, $$\operatorname{Var}[W] = \operatorname{Var}[D_1 + \cdots + D_7] \overset{\text{ind}}{=} \sum_{i=1}^7 \operatorname{Var}[D_i] = 7\operatorname{Var}[D] = 7 \sigma_D^2,$$ when each daily forecast in the week is independent of the others. This is what is going on in your question. The formula $$\operatorname{Var}[aX] = a^2 \operatorname{Var}[X]$$ does not apply here.

• Thank you. This was such an intuitive answer :) It clarified my doubts. – Anirudh Murali Dec 26 '17 at 6:08