query: If $E(XY)=E(X) \cdot E(Y)$, why is not $E(X^2)=[E(X)^2]$
positive_passages.docid: 27788
positive_passages.text: Why is the expected value $E(X^2) \neq E(X)^2$?
negative_passages.docid: 45143
negative_passages.text: Identify $E(X\mid X+Y)$ for $X$ and $Y$ independent and Poisson
negative_passages.docid: 35454
negative_passages.text: Is it correct to write $\Bbb E[X]$ or $\Bbb E_{\theta}[X]$?
negative_passages.docid: 62605
negative_passages.text: How to find $\mathbb{E}(X\mid\mathbf{1}_{X<Y})$ where $X,Y$ are i.i.d exponential variables?
negative_passages.docid: 95908
negative_passages.text: What is the meaning of $\exp(\,\cdot\,)$?
negative_passages.docid: 100968
negative_passages.text: Why is $|e^{i \lambda z}| |e^{- \lambda y}|= |e^{- \lambda y}|$ here?
negative_passages.docid: 125209
negative_passages.text: Prove $e^x - e^y \leq e |x-y|$ for $x$ belonging to $[0,1]$
negative_passages.docid: 48270
negative_passages.text: How to prove that the random variables $X$ and $X^2$ are not independent?
negative_passages.docid: 122719
negative_passages.text: Show that random variables $X$ and $Y$ are not independent, but nevertheless Cov$[X,Y] = 0$
negative_passages.docid: 85818
negative_passages.text: $E(|X-Y|)$ for $X$, $Y$ i.i.d and uniform on $[0,1]$
negative_passages.docid: 111088
negative_passages.text: Conditional probability, $E[E(X\mid Y)\mid Y,Z]=?$
negative_passages.docid: 662
negative_passages.text: If $E[X|Y]=Y$ almost surely and $E[Y|X]=X$ almost surely then $X=Y$ almost surely
negative_passages.docid: 101394
negative_passages.text: Please explain the logic behind $d(xy) = y(dx) + x(dy)$
negative_passages.docid: 91216
negative_passages.text: Question about Conditional Expectation
negative_passages.docid: 128764
negative_passages.text: How does $y[x] = a x^r$ when $y'[x] = \frac{(r y[x])}{x}$ and $y[1] = a$
negative_passages.docid: 5350
negative_passages.text: Explain why $E(|X|^p)=p\int_{0}^{\infty}y^{p-1}P(|X|>y) dy$
negative_passages.docid: 34693
negative_passages.text: Prove $E^{\mathbb Q}[Y]=E^{\mathbb P}[XY]$ if $E^{\mathbb P}[X]=\mathbb P(X>0)=1$ and $ \mathbb Q(A)=E^{\mathbb Q}[X1_A] $