The mean square deviations of a set of observations x1, x2, ..., xn about a point c is defined to be

Q: Why do we treat $c$ as a variable? Isn't it just a constant?

In the definition of MSD, $c$ is a parameter. By treating it as a variable, we reveal the parabolic nature of the deviation, which is a powerful tool in statistics. It allows us to see that the MSD is minimized at the mean, $\bar{x}$.

Q: Can we solve this without expanding the summation?

You could, but expansion is the most reliable way to isolate the mean $\bar{x}$ and the mean of squares, which are the building blocks of variance. It transforms a complex statistical expression into a simple system of linear equations.

Q: Why is the minimum of the MSD parabola at the mean?

If you take the derivative of $f(c) = \frac{1}{n} \sum (x_i - c)^2$ with respect to $c$ and set it to zero, you get $\frac{2}{n} \sum (x_i - c) = 0$, which simplifies to $\sum x_i - nc = 0$, or $c = \bar{x}$. This confirms that the vertex of the parabola is indeed at the arithmetic mean.

Question

The mean square deviations of a set of observations $x_1, x_2, \dots, x_n$ about a point $c$ is defined to be $\frac{1}{n} \sum_{i=1}^n (x_i - c)^2$. The mean square deviations about $-1$ and $+1$ of a set of observations are 7 and 3 respectively. Find the standard deviation of this set of observations.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plain, and you are tasked with finding a single point that best represents a set of data points $x_1, x_2, \dots, x_n$. You want to minimize the 'spread' or 'deviation' of these points from your chosen point $c$. This is the essence of Mean Square Deviation (MSD).

When we define the function:
$$MSD(c) = \frac{1}{n} \sum_{i=1}^n (x_i - c)^2$$
we are describing a beautiful, upward-facing parabola. If you expand $(x_i - c)^2$, you get $x_i^2 - 2cx_i + c^2$.

When you sum this over all $n$ points and divide by $n$, you get a quadratic expression in terms of $c$. Every quadratic function $f(c) = Ac^2 + Bc + C$ is a parabola. The lowest point, or the vertex, occurs exactly at the arithmetic mean, $\bar{x}$.

### Translating the Problem into Algebra

The problem provides two specific points on this parabola. We are told the MSD about $c = -1$ is $7$, and the MSD about $c = 1$ is $3$.

For $c = -1$, we have:
$$\frac{1}{n} \sum_{i=1}^n (x_i + 1)^2 = 7$$
Expanding this, we get $\frac{1}{n} \sum x_i^2 + 2\bar{x} + 1 = 7$. Let us call this **Equation (1)**.

For $c = 1$, we have:
$$\frac{1}{n} \sum_{i=1}^n (x_i - 1)^2 = 3$$
Expanding this, we get $\frac{1}{n} \sum x_i^2 - 2\bar{x} + 1 = 3$. This is **Equation (2)**.

### The Elegant Cancellation

We now have a system of two linear equations with two unknowns: the mean of squares $\frac{1}{n} \sum x_i^2$ and the mean $\bar{x}$.

If we subtract Equation (2) from Equation (1), the mean of squares and the constant $1$ vanish:
$$(2\bar{x}) - (-2\bar{x}) = 7 - 3$$
$$4\bar{x} = 4 \Rightarrow \bar{x} = 1$$

To find the mean of squares, we add the two equations:
$$2\left(\frac{1}{n} \sum x_i^2 + 1\right) = 10$$
$$\frac{1}{n} \sum x_i^2 + 1 = 5 \Rightarrow \frac{1}{n} \sum x_i^2 = 4$$

### Connecting to Variance and the Final Answer

The variance $\sigma^2$ is defined as the mean of the squares minus the square of the mean:
$$\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$$

Substituting our calculated values:
$$\sigma^2 = 4 - (1)^2 = 3$$

Geometrically, this variance is the minimum value of our MSD parabola. Since the mean is $1$, and we were given that the MSD at $c = 1$ is $3$, we have confirmed that the vertex of our parabola is at $c = 1$ with a minimum value of $3$.

Finally, the standard deviation $\sigma$ is the square root of the variance:
$$\sigma = \sqrt{3}$$

The final answer is **$\sqrt{3}$**.

Visualized Solution

Understanding Mean Square Deviation

MSD about $c = - 1$

Expanding the First Equation

MSD about $c = 1$

Expanding the Second Equation

Subtracting the Equations

Adding the Equations

The Variance Formula

Calculating Variance

Final Result: Standard Deviation

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Translating the Problem into Algebra

The Elegant Cancellation

Connecting to Variance and the Final Answer

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