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Decomposition of variance

In the case where $X$ is symmetric about zero (as in your illustration, though not in the conditions of the problem), with mass function $f(x)$ such that $f(-x)=f(x)$, $A$ has mass function $2f(x)\; [...
mcd's user avatar
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4 votes
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Is Cov (X,XY) positive if X,Y >0

No. Take $X$ a positive random variable for which $\mathbb{E}[X]$ and $\mathbb{E}[1/X]$ both exist, then take $Y = 1/X^2$. Observe that, by Jensen's inequality, $\mathbb{E}[1/X] \geq 1/\mathbb{E}[X]$, ...
Julius's user avatar
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1 vote

Efficient and unbiased estimation of the location ($\mu$) of truncated normal distribution with known scale ($\sigma^2$) and truncation points

One observation is not enough to infer $\mu$ with much accuracy at all. I think you are inherently going to be limited by your lack of data. I suspect the best you can do is take a Bayesian approach. ...
D.W.'s user avatar
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the first principal component’s variance

PCA is based on the covariance matrix $S = X^T X / N$. In (3.48) the division by $N$ was omitted and in (3.49) it is again accounted for.
Markéta Makarová's user avatar
0 votes
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Variance of a Continuous Stochastic Process

Here is a less rigorous but nevertheless justifiable computation for the mean and variance of $\gamma_t$. We start from the SDE $$ \gamma_{t+\mathrm{d}t} - \gamma_t = \mathrm{d}\gamma_t = -\lambda(\...
Sangchul Lee's user avatar
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The variance of the resulting Gaussian is found by taking the second derivative of the exponent?

I've read the paper recently and wondered the same thing. The Cramér–Rao bound (see this and this) states that the variance of an unbiased estimator cannot be lower than the inverse Fisher information,...
Ivan Bilić's user avatar
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What is the value of $\operatorname{Var}(S^2)$, when $S^2$ is given as sample variance with denominator $n$ instead of $(n-1)$?

I will assume that $X_i$'s are IID, having finite 4th moment. Note that \begin{align*} S^2 = \frac{1}{n} \sum_{i=1}^{n} (X_i - \overline{X})^2 = \frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2 - (\overline{X}...
Sangchul Lee's user avatar

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