1- What does the variance of expected value tell us?
2- Why when calculating the variance of expected value we don’t divide it by n?
1- What does the variance of expected value tell us?
2- Why when calculating the variance of expected value we don’t divide it by n?
informally measures how far a set of (random) numbers are spread out from their mean. (Mean is the expected value).
While using the Expected Value - you have probability of occurrence of each value. Since probability is available - the n has already been incorporated in that value.
e.g. with a probability of 15% your value could be 4, 45% it could be 6 and 40% it could be 8… so you have the sum of probabilities = 1 and you then use these to calculate the expected value as 6.5 …
Think of #2 this way: when you add 1, 2, 3, and 4, and then divide the sum by 4 (for a normal arithmetic mean), in essence you’re multiplying 1 by ¼, 2 by ¼, 3 by ¼, and 4 by ¼, then adding those products; note that the sum ¼ + ¼ + ¼ + ¼ = 1, so you can think of each ¼ as a weight or probability for its corresponding value (1, 2, 3, or 4). When computing expected value, you’re doing the same thing; the “dividing by n” is already incorporated.
Thank you very much.
My pleasure.
Am I missing something here with this idea of “variance of expected value”? This is the second thread about it…An expected value (like mu) is a constant (zero variance).