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31. My proof is similar to @Math1000's but is useful if you're not familiar with generating functions. However, I'm using the variant of the geometric distribution where X is the number of trials until success. Therefore E[X] = 1 p. (Both variants have the same variance). We know E[X] = 1 p. Then the variance is: E[X2] − (E[X])2 = E[X(X − 1 ...
2. If X and Y are random variables with correlation coefficient 0.7, each of which has variance 6, what is the variance of X−Y? Enter your answer as a decimal. Using the information given, I was able to determine the Covariance of X and Y to be 4.2. I thought maybe the variance of X-Y would be 0 but that's too easy.
I was asked to derive the mean and variance for the negative binomial using the moment generating function of the negative binomial. However i am not sure how to go about using the formula to go out and actually solve for the mean and variance.
9. There can be some confusion in defining the sample variance ... 1/n vs 1/ (n-1). The OP here is, I take it, using the sample variance with 1/ (n-1) ... namely the unbiased estimator of the population variance, otherwise known as the second h-statistic: h2 = HStatistic[2][[2]] These sorts of problems can now be solved by computer.
$\begingroup$ Funny thing is that given the density of Gaussian you do not need even an integration to find the mean and variance! $\endgroup$ – Arash Commented Oct 8, 2013 at 0:40
Find the expectation and variance of X. 4. Finding out the Probability Mass Function from a Probability ...
Is there a formula for the variance of a (continuous, non-negative) random variable in terms of its CDF? The only place I saw such formula was is Wikipedia's page for the Variance (https://en.wiki...
Calculating the variance can be done using Var(X) =E(X2) −E(X)2. E(X2) = ∫x2f(x)dx = 47 24. So the variance is equal to: Var(X) = 47 24 −(31 24)2 ≈ 0.29. Just used what you had previously posted to figure it out - was about to type my solution to make sure I had it when I saw your edit and I've got it.
V(ˉX) = V(1 nT) = (1 n)2V(T) = (1 n)2nσ2 = 1 nσ2 = σ2 / n. Notes: (1) In the first displayed equation the expected value of a sum of random variables is the sum of the expected values, whether nor not the random variables are independent. (2) However, the variance of the sum of random variables is not necessarily equal to the sum of the ...
40. The mean absolute deviation is: ∑n i=1|xi −x¯| n. The variance is: ∑n i=1(xi −x¯)2 n − 1. So the mean deviation and the variance are measuring the same thing, yet variance requires squaring the difference. Why? Squaring always gives a non-negative value, but the absolute value is also a non-negative value.