How do you find the variance of Xi?
Each Xi has variance p × (1 − p)2 + (1 − p) × (0 − p)2 = p(1 − p). By independence of the Xi , the sum of the Xi has variance var(X1) + + var(Xn) = np(1 − p). Notice that the independence between the Xi is not used for the calculation of the mean, but it is used for the calculation of the variance.
What is XI in mean formula?
xi represents the ith value of variable X. For the data, x1 = 21, x2 = 42, and so on. For the data, Σxi = 21 + 42 +… + 52 = 290.
Is the variance of the sum of independent random variables 0?
We start by expanding the definition of variance: Now, note that the random variables and are independent, so: But using (2) again: is obviously just , therefore the above reduces to 0. So, coming back to the long expression for the variance of sums, the last term is 0, and we have:
How to calculate the variance of a function?
For a function : The variance of X is defined in terms of the expected value as: From this we can also obtain: Which is more convenient to use in some calculations. From the definitions given above it can be easily shown that given a linear function of a random variable: , the expected value and variance of Y are:
How to calculate the expected value of a random variable?
Formally, the expected value of a (discrete) random variable X is defined by: Where is the PMF of X, . For a function : The variance of X is defined in terms of the expected value as: From this we can also obtain: Which is more convenient to use in some calculations.
Which is the last term of the variance of sums?
So, coming back to the long expression for the variance of sums, the last term is 0, and we have: As I’ve mentioned before, proving this for the sum of two variables suffices, because the proof for N variables is a simple mathematical extension, and can be intuitively understood by means of a “mental induction”. Therefore: