Given that $\bar{x}$ is the mean and $\sigma^2$ is the variance of $n$ observations $x_1, x_2, \ldots, x_n$. Prove that the mean and variance of the observations $ax_1, ax_2, ax_3, \ldots, ax_n$ are $a\bar{x}$ and $a^2\sigma^2$, respectively, $(a
eq 0)$.

Question

Given that $\bar{x}$ is the mean and $\sigma^2$ is the variance of $n$ observations $x_1, x_2, \ldots, x_n$. Prove that the mean and variance of the observations $ax_1, ax_2, ax_3, \ldots, ax_n$ are $a\bar{x}$ and $a^2\sigma^2$, respectively, $(a 
eq 0)$.

Accepted Answer

> [!TLDR]
> Scaling every observation by a constant $a$ scales the mean by $a$ and the variance by $a^2$, because mean is a linear operator and variance measures squared deviations.

## Why This Works — The Intuition

When you multiply every data point by a constant $a$, two things happen:

- The **centre** of the data shifts proportionally — if the original average was $\bar{x}$, the new average is simply $a$ times that.
- The **spread** changes in a squared sense. Variance measures average squared distance from the mean. If each point and the mean both get multiplied by $a$, each deviation gets multiplied by $a$, so each *squared* deviation gets multiplied by $a^2$. The average of those squared deviations therefore also gets multiplied by $a^2$.

This is a fundamental property: **mean scales linearly, variance scales quadratically**. Let's prove it cleanly.

---

## Step-by-Step Proof

### 1. Define the original mean and variance

The mean of the original observations is:

$$
\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i
$$

The variance is:

$$
\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2
$$

> [!NOTE]
> We are using the population variance formula (division by $n$), which is standard in this context. If you use $n-1$ for sample variance, the same scaling property holds — the constant $a$ still factors out as $a^2$.

---

### 2. Find the mean of the scaled observations

Let the new observations be $y_i = a x_i$ for $i = 1, 2, \ldots, n$.

Their mean is:

$$
\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{1}{n} \sum_{i=1}^{n} a x_i = a \cdot \frac{1}{n} \sum_{i=1}^{n} x_i = a \bar{x}
$$

So the new mean is $a\bar{x}$. That's the first result.

> [!TIP]
> This works because summation and multiplication by a constant commute — you can pull $a$ out of the sum. Mean is a **linear** function of the data.

---

### 3. Find the variance of the scaled observations

The variance of the $y_i$ is:

$$
	ext{Var}(y) = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2
$$

Su

Q.Given that $\bar{x}$ is the mean and $\sigma^2$ is the variance of $n$ observations $x_1, x_2, \ldots, x_n$ . Prove that the mean and variance of the observations $ax_1, ax_2, ax_3, \ldots, ax_n$ are $a\bar{x}$ and $a^2\sigma^2$ , respectively, $(a \neq 0)$ .

Concept understanding — Effect of Scaling Variance

Effect of Scaling Variance – First Encounter

Intuition first

The precise statement

What about adding a constant?

A quick example

Common mistake to avoid

Why This Works — The Intuition

Step-by-Step Proof

1. Define the original mean and variance

2. Find the mean of the scaled observations

3. Find the variance of the scaled observations

Why This Works — The Intuition

Step-by-Step Proof

1. Define the original mean and variance

2. Find the mean of the scaled observations

3. Find the variance of the scaled observations

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