24 votes
Accepted

Expectation of the product of iid random variables

First, let's establish the correct identity. When $X_1, \ldots, X_N$ are independent variables with finite expectations $\mu_i=X_i,$ then by laws of conditional expectation, $$E\left[\prod_{i=1}^N X_i\...
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10 votes

Expectation of the product of iid random variables

As an addendum to @whuber's explanation, consider that \begin{align} \prod_{i=1}^N \vert X_i\vert &= \exp\{ \sum\nolimits_{i=1}^N \overbrace{\ln \vert X_i\vert}^{Y_i} \}\\ &= \exp\{ \sum\...
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7 votes

Expectation of the product of iid random variables

There are several things to note here. Multiplying lots of numbers in computer leads to rounding errors, especially if these numbers vary in magnitude, i.e. small numbers are multiplied by large ...
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6 votes

Why do confidence intervals width increase when sample size increases "a bit"?

A standard confidence interval for a mean is calculated using $s/\sqrt n$. If you increase $n$ but also increase the sample standard deviation $s$ by enough to offset the larger sample size, then your ...
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5 votes
Accepted

Upper bound for random correlation

As discussed here, the sampling distribution of correlation for uncorrelated Gaussian variables covers the entire range of $[-1,1]$. While you have not assumed Gaussian variables, you have not ruled ...
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5 votes

Can spurious correlations exist in the (theoretical) population?

With spurious correlation I assume that you mean the description on the tag Nonzero correlation between variables $𝑥$ and $𝑦$ where neither $𝑥$ causes $𝑦$ nor $𝑦$ causes $𝑥$. May result from ...
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5 votes

Can spurious correlations exist in the (theoretical) population?

As suggested in Ben's answer in the thread "Spurious relationships: flavours, terminology", the qualifier spurious can be attributed to our interpretation of some statistical finding. We may ...
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3 votes
Accepted

On Probability Space and Random Variables

Henry's answer is good. For the first question, I wanted to point you to the Kolmogorov extension theorem, which gives sufficient conditions under which you can put a collection of random variables ...
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2 votes
Accepted

What transformation(s) $g$ satisfy $\text{Cov}(g(X), g(Y)) \le c(g) \text{Cov}(X,Y)$?

Consider, as you say, standard Normal variates and $g(x) = x/10$ if $|x| < 100$, $0$ otherwise. It should be intuitively obvious that the covariance between $g(x)$ and $g(y)$ is less than that of $...
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1 vote

On Definition of Random Variables

You can write the desired set using only negation and countable unions of the foundational sets as: $$\begin{align} \{ a \leq X < b \} &= \{ a \leq X \leq b \} - \overline{ \{ X \leq b \}} \\...
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1 vote

On Definition of Random Variables

To quote from Wikipedia, a $\sigma$-algebra is defined as $\Omega$ is in $\mathcal F$, and $\Omega$ is considered to be the universal set in the following context. $\mathcal F$ is closed under ...
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1 vote
Accepted

On Definition of Random Variables

As @whoknowsnot has shown, for every $a \in \mathbb R$, the event $\{X< a\}\in \mathcal F$, and since $\{X\leq a\}\in \mathcal F$ by definition, the difference of these two events, namely $\{X = ...
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1 vote

Can two random variables, both of which are dependent on a 3rd random variable, be independent of each other?

This is possible. Two examples: Discrete example Throw two coins independently, but code the outcomes as 0 or 1. Then define: $$ \begin{align} X&=\text{outcome on first coin}\\ Y&...
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