Discrete random variance transformation

For a discrete random variable, called X, if we know that there is another DRV which we can write as Y=aX+b - where a and b are constants:

Var(Y)=a^2\,Var(X)

We ignore the added constant and multiply the old variance by the square of a.

Why square a when finding the variance?

If we didn’t square a, we’d be finding the standard deviation (if multiplying a\times\sigma). If we want to find the variance from the standard deviation, then we’ll need to square the result, so (a\sigma)^2=a^2\,Var(X)

Why do we ignore the constant?

If we add a constant (b) to a discrete random variable, it shifts the distribution one way, but doesn’t actually change how far apart the values are. That means the variance

flashcards

QuestionAnswer
What is the formula for variance of a transformed discrete random variable Y=aX+b?Var(Y)=a^{2}\,Var(X)
What happens to variance when a constant b is added to a discrete random variable?The variance is unchanged because the constant shifts the distribution but does not change the spread of values.
Why is the constant b ignored in Var(aX+b)=a^{2}Var(X)?Adding a constant shifts the distribution but does not affect how far apart the values are, so the variance remains the same.
Why is a squared in Var(aX+b)=a^{2}Var(X)?If a were not squared, the calculation would find the standard deviation (a\sigma). To find variance from standard deviation, the result must be squared, giving (a\sigma)^{2}=a^{2}Var(X).