How does the variance of a temperature reading change when you convert Celsius to Fahrenheit?

## The rule For a random variable X and constants a and b: - `E[aX + b] = a*E[X] + b` - `Var(aX + b) = a^2 * Var(X)` - `sd(aX + b) = |a| * sd(X)` The mean follows the transformation exactly; the spread ignores the shift and squares the scale factor. ## Why the shift drops out Variance is defined from deviations around the mean: `Var(X) = E[(X - E[X])^2]`. Let `Y = X + b`. Then `E[Y] = E[X] + b`, so the deviation is `Y - E[Y] = (X + b) - (E[X] + b) = X - E[X]` The b cancels. Every observation and the mean moved by the same amount, so no distance between them changed. Spread is a statement about relative position, and a rigid translation preserves all relative positions. ## Why the scale factor is squared Let `Y = aX`. Then `E[Y] = a*E[X]` and `Y - E[Y] = a*(X - E[X])` Squaring gives `a^2 * (X - E[X])^2`, and taking expectations pulls the constant `a^2` out: `Var(aX) = a^2 * Var(X)` The square is why the sign of a is irrelevant: `Var(-X) = Var(X)`. Flipping a variable around leaves its spread alone. It is also why variance carries squared units: doubling a measurement in seconds quadruples a variance measured in seconds squared. Taking the square root returns to the original units and to a linear scale factor, `|a|`. ## The Celsius-to-Fahrenheit case `F = 1.8*C + 32`. Suppose a sensor reports Celsius readings with mean 20 and standard deviation 2, hence variance 4. - Mean: `E[F] = 1.8*20 + 32 = 68` degrees Fahrenheit. - Variance: `Var(F) = 1.8^2 * 4 = 3.24 * 4 = 12.96` degrees Fahrenheit squared. - Standard deviation: `sd(F) = 1.8 * 2 = 3.6` degrees Fahrenheit, which is `sqrt(12.96)` as it must be. Two mistakes appear over and over. The first is adding 32 to the spread, producing an absurd standard deviation of 33.8 degrees for a sensor that is accurate to a couple of degrees. The second is scaling the variance by 1.8 rather than 3.24, which quietly understates the spread by nearly half. The same arithmetic covers any unit change: cents to dollars divides the variance by 10,000 and the standard deviation by 100; milliseconds to seconds divides the variance by a million. Because the coefficient of variation, the standard deviation divided by the mean, is dimensionless, it is unchanged by a pure rescaling with `b = 0`, though the plus-32 shift in the Fahrenheit case does change it. ## Standardising The rule is what makes standardisation work. Define `Z = (X - mu)/sigma`, where mu and sigma are the mean and standard deviation of X. This is affine with `a = 1/sigma` and `b = -mu/sigma`, so - `E[Z] = (E[X] - mu)/sigma = 0` - `Var(Z) = (1/sigma)^2 * sigma^2 = 1` Mean zero, variance one, whatever units X started in. Standardisation does not change the shape of the distribution, only its location and scale. ## Differences and the sign trap A closely related question: what is the variance of a difference of two independent variables? Because the coefficient on the second is -1 and it gets squared, `Var(X - Y) = Var(X) + Var(Y)` for independent X and Y The variances add, they do not subtract. Uncertainties compound whether you are adding two quantities or subtracting them, which is why the spread of a difference of two independent measurements is wider than either one. Candidates who answer `Var(X) - Var(Y)` have carried the minus sign through a step where it must be squared, and in the case of equal variances that error produces zero spread for a difference that is obviously variable. The independence assumption matters here: when the two quantities co-move, the variance of their sum or difference is not simply the sum of the parts. ## Quick self-check If someone tells you a transformation changed the variance but not the mean, or changed the mean by a factor and the variance by the same factor, something is wrong. Under an affine map the mean picks up both a and b; the variance picks up `a^2` and ignores b entirely.