How can I calculate the potential values of the final value from the value of standard deviation?

Question

If I have a list of 99 values, say $x_1$ to $x_{99}$, and I have a value for the standard deviation of an $x_{1}$ to $x_{100}$ data set, containing these 99 values + 1 other, can I calculate the potential values for the extra data point? Is there a simple algebraic solution to this that I am missing?

Answer 1

We are basically comparing two data sets, one with $99$ entries and another with $100$ entries. In the latter, the first $99$ are the same as in the first set and the last value is "unknown". Additionally, we know

• the mean of the first $99$ values, $\mu_{99}$
• the variance of smaller set, $\sigma^2_{99}$
• the variance of larger set, $\sigma^2_{100}$

What we want to know, is the value of the last entry on the larger set, the unknown value $x_{100}$. We can start by considering how to calculate means and variances. It's easy to see that the mean of the larger set is just $$\tag{1} \mu_{100} = \frac{99 \mu_{99}+x_{100}}{100} $$ The variances are $$\tag{2} \sigma^2_{99} = \frac{1}{99}\sum_{i=1}^{99} \left(x_i - \mu_{99} \right)^2 $$ $$\tag{3} \sigma^2_{100} = \frac{1}{100}\sum_{i=1}^{100} \left(x_i - \mu_{100} \right)^2 $$ Let's try to write the latter in terms of the former. Splitting the sum and using Equation ($1$), we get $$ \begin{split} \sigma^2_{100} =& \frac{1}{100}\sum_{i=1}^{99} \left(x_i - \mu_{100} \right)^2 + \frac{1}{100} \left(x_{100} - \mu_{100} \right)^2\\ =& \frac{1}{100}\sum_{i=1}^{99} \left(x_i - \frac{99 \mu_{99}+x_{100}}{100} \right)^2 + \frac{1}{100} \left(x_{100} - \frac{99 \mu_{99}+x_{100}}{100} \right)^2\\ =& \frac{1}{100}\sum_{i=1}^{99} \left(\frac{100x_i -99 \mu_{99}-x_{100}}{100} \right)^2 + \frac{1}{100} \left(\frac{100x_{100}-99 \mu_{99}-x_{100}}{100} \right)^2\\ =& \frac{1}{100}\sum_{i=1}^{99} \left(\frac{100x_i -100 \mu_{99} + \mu_{99}-x_{100}}{100} \right)^2 + \frac{1}{100} \left(\frac{99x_{100}-99 \mu_{99}}{100} \right)^2\\ =& \frac{1}{100}\sum_{i=1}^{99} \left([x_i -\mu_{99}]+\left[\frac{\mu_{99}-x_{100}}{100}\right] \right)^2 + \frac{99^2}{100} \left(\frac{x_{100}- \mu_{99}}{100} \right)^2\\ \end{split} $$ Let's look at the first term. We have $$ \begin{split} \sum_{i=1}^{99} \left([x_i -\mu_{99}]+\left[\frac{\mu_{99}-x_{100}}{100}\right] \right)^2 =& \sum_{i=1}^{99} \left\{ [x_i -\mu_{99}]^2 + 2[x_i -\mu_{99}]\left[\frac{\mu_{99}-x_{100}}{100}\right] + \left[\frac{\mu_{99}-x_{100}}{100}\right]^2 \right\} \\ =& \underbrace{\sum_{i=1}^{99} [x_i -\mu_{99}]^2}_{=99\cdot \sigma^2_{99}} + 2\underbrace{\sum_{i=1}^{99}[x_i -\mu_{99}]\left[\frac{\mu_{99}-x_{100}}{100}\right]}_{=0} + \underbrace{\sum_{i=1}^{99}\left[\frac{\mu_{99}-x_{100}}{100}\right]^2}_{=99\cdot \left[\frac{\mu_{99}-x_{100}}{100}\right]^2} \\ =& 99 \sigma^2_{99} + 99\cdot \left[\frac{\mu_{99}-x_{100}}{100}\right]^2 \end{split} $$ Plugging this back in, we have the equation $$ \sigma^2_{100} = \frac{99}{100} \sigma^2_{99} + \frac{99}{100}\cdot \left(\frac{x_{100}- \mu_{99}}{100}\right)^2 + \frac{99^2}{100} \left(\frac{x_{100}- \mu_{99}}{100} \right)^2 $$ Phew. Can you continue from here?