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to translations along the $x$ axis: the central moments of $Y = X + a$ are the same as the central moments of $X$. … From a computational perspective, in some cases, (non-central) moments are slightly easier to calculate than central moments, while in other cases, the central moments are slightly easier to calculate …

An alternative viewpoint is that $X$ has a mixture density that consists of an equally-weighted sum of the density of an exponential random variable with parameter $2$ (the $e^{-2x}\mathbf 1_{\{x>0\} …

The complementary CDF of the random variable $X$ with hazard rate $\lambda(t)$ is $$1- F_X(x) = \exp\left(-\int_0^x \lambda(t) \,\mathrm dt\right).$$ The complementary CDF of the random variable $Y$ w …

I'm interested in the width of the distribution of $Y_m$. How does its variance scale with the sample size $N$? $\text{var}(X_i^m) = E[X_i^{2m}] - (E[X_i^m])^2$ is easily evaluated from the momen …

Just expand out the right side using FOIL to get $$\begin{align} \sum_{i=1}^n (X_i - \bar{X})^2 &= \sum_{i=1}^n (X_i^2 - 2X_i\bar{X} + \bar{X}^2)\\ &= \left(\sum_{i=1}^n X_i^2\right) -2\bar{X}\left(\s …

For convenience, let $X$ denote a continuous zero-mean random variable with density function $f(x)$, and consider $P\{X \geq a\}$ where $a > 0$. We have $$P\{X \geq a\} = \int_a^{\infty}f(x)\,\mathrm …

Hint: The integrand $e^{(t-1)x}$ has value $1$ if $t=1$ and is an increasing function of $x$ if $t > 1$. What do you suppose is the value of the integral when $t \geq 1$? In particular, is the value a …

Note that $S^2$ has terms involving $X_i^4$, and so $E[S^2]$ is the sum of terms involving $E[X_i^4]$. Thus, if the fourth moment is not finite, neither is $E[S^2]$ finite, nor is var$(S^2)$ finite. …

already found that $$M_X(t) = \frac{1}{1-t^2} = 1 + t^2 + t^4 + \cdots + t^{2n} + \cdots \tag{2}$$ (that's the "by hook or by crook" part) and so he can compare $(1)$ and $(2)$ to figure out what the moments …