statistics - Prove variance in Uniform distribution (continuous ...
Mar 26, 2014 · Add a comment. 3 Answers. Sorted by: 35. By the definition of the variance, Var X =E[X2] − (EX)2 Var. . X = E [X 2] − (E X) 2. Since here EX = 1 b−a ∫[a,b] xdx = a+b 2 E X = 1 b − a ∫ [a, b] x d x = a + b 2, and EX2 = 1 b−a ∫[a,b] x2dx = b3−a3 3(b−a) = a2+ab+b2 3 E X 2 = 1 b − a ∫ [a, b] x 2 d x = b 3 − a 3 3 (b − ...
Derivation of Variance of Discrete Uniform Distribution over …
I'm trying to prove that the variance of a discrete uniform distribution is equal to $\cfrac{(b-a+1)^2-1}{12}$. I've looked at other proofs, and it makes sense to me that in the case where the distribution starts at 1 and goes to n, the variance is equal to $\cfrac{(n)^2-1}{12}$.
$U[a,b]$ has maximum variance of any distribtuion on $[a,b]$
Mar 5, 2015 · Because ${\sf Var}[X] = {\sf E}[(X-\mu)^2]$ , therefore any distribution on the support $[a;b]$ is going to have greater variance than a uniform distribution whenever points nearer the bounds have higher density than points nearer the mean. $\endgroup$ –
Finding the method of moments estimator for the Uniform …
The first moment is $$ \int_a^b x f(x)\,dx = \int_a^b \frac{x\,dx}{b-a} = \frac 1 2 \cdot \frac{b^2-a^2}{b-a} = \frac{b+a} 2. $$ The second moment is $$ \int_a^b x^2 f(x) \,dx = \int_a^b \frac{x^2\,dx}{b-a} = \frac 1 3 \cdot \frac{b^3 -a^3}{b-a} = \frac{b^2+ba+a^2} 3. $$ So equate the sample moments with the population moments found above: \begin{align} & …
probability - uniform distribution mean and variance
Jun 13, 2017 · Sorted by: 1. Given that X X has a discrete uniform distribution that can have values from −2 − 2 to +2 + 2, calculate. This is unclear, but if you interpret it to mean the integer values from −2 − 2 to 2 2, that is X ∼ [−2; 2] ∩N X ∼ [− 2; 2] ∩ N, then indeed... A)the number of possible values of X, n X, n. We have n ...
statistics - Unbiased estimator of a uniform distribution
Apr 22, 2017 · 6. To show that the sample maximum xmax = maxni = 1{xi} is an unbiased estimator of θ you would need to show that E(xmax) = θ. This is saying that the average value of the maximum of n uniform variables on [0, θ] is θ. This cannot be right. θ is the maximum value that any of the uniform variables can have. xmax, the sample maximum might ...
Variance for uniform distribution - Mathematics Stack Exchange
Jul 26, 2017 · This question seems to be more about generic properties of the variance rather than about the uniform distribution itself. The mapping (X, Y) ↦ Cov(X, Y) (covariance of X and Y) is bilinear. The variance, X ↦ Var(X) = Cov(X, X) is the quadratic form associated to Cov. This is why, when you expand Var(aX + bY), you obtain: a2Var(X) + b2Var(Y ...
How do I calculate the mean and variance of a uniform …
Sep 21, 2021 · Random variable X with uniform distribution over the interval (2,7) Find the moment generating function and use it to obtain the mean and variance of X. I was able to find the mgf which is given by: (e^7t - e^2t)/5t and the first derivative of the mgf, but division by 0 occurs when I substitute t = 0.
probability - Expectation, variance etc for uniform distribution ...
Dec 1, 2017 · Expectation, variance etc. for exponential distribution 0 Finding the Expectation and Variance, given the distribution function and density function for a continuous random variable
Distribution of the maximum of $n$ uniform random variables
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