4

$X$ and $Y^X$ are not independent variables in general. $Y^X$ is a function of the variable $X$ and thus depends on the distribution of $X$. Another way to think about independence is about information, knowing the value of $Y^X$ would give you information about the value of $X$. Re: misconception. The way you have written it, $X$ in $Y^X$ is the same random ...


3

It is well known that the mean E(Y |X) minimizes the Root Mean Square Error (RMSE). The use of the term 'error' is confusing in this context. We could better specify this more precisely and say that the sample mean (that is different from $E(Y|X)$) minimizes the root mean square of residuals. (Related question: Difference between mean square residual and ...


3

Let's solve the more general problem and then apply it to the specific setting as an illustration. Suppose $g:\mathcal{X}\to\mathbb{R}$ is a measurable function. Let $\{\mathcal{A}_i\}\subset \mathscr{P}(\mathcal{X})$ be a finite or countable collection of subsets of $\mathcal X,$ each with finite positive measure $p_i = \int_{\mathcal{A}_i}\mathrm{d}x.$ ...


2

What you are looking for is the expected profit from the bet. It's denoted $E(X)$ where $X$ is a "placeholder" for the random profit you would get from a game. Notice that the profit can also be negative. For the bet on red your expected profit is, as you correctly calculated, \begin{align} E(X) &= P(X=-1)\times(-1) + P(X=1)\times1\\ &=...


2

That expectation would be $\infty$, as we can interpret, in this context, $1/0$ as infinity. One reason is that the post mentioned, as an example, the binomial distribution, which is nonnegative. So this solution can only be defended, if at all, for nonnegative random variables. The principled solution is in the next paragraph. See also I've heard that ...


1

It'd be simplest to prove $\operatorname{cov}(X,Y+Z)=\operatorname{cov}(X,Y)+\operatorname{cov}(X,Z)$ first and go from there. $$\begin{align}\operatorname{cov}(X,Y+Z)&=\mathbb{E}[X(Y+Z)]-\mathbb E[X]\mathbb E[Y+Z]\\&=\mathbb E[XY]+\mathbb E[XZ]-\mathbb E[X]\mathbb E[Y]-\mathbb E[X]\mathbb E[Z]\\&=(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])+(\mathbb ...


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