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seen Aug 23 at 1:35

"The quieter you become, the more you are able to hear..."


Aug
21
awarded  Popular Question
Jul
29
comment Independent of random variables.
@StefanHansen thanks very much.
Jul
21
asked conditional probability about sum and product rule
Jul
13
accepted Terminology on algebra.
Jul
12
asked Terminology on algebra.
Jul
9
awarded  Civic Duty
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jul
2
asked Independent of random variables.
Jun
9
asked The maximum term in Binomial distribution
Jun
7
accepted Asymptotic in hypergeometric distribution.
Jun
6
asked Asymptotic in hypergeometric distribution.
May
31
accepted Does the inequality $\frac{1-x_1x_2}{1-y_1y_2} \leq \frac{1-x_1}{1-y_1} + \frac{1-x_2}{1-y_2}$ hold?
May
31
asked Does the inequality $\frac{1-x_1x_2}{1-y_1y_2} \leq \frac{1-x_1}{1-y_1} + \frac{1-x_2}{1-y_2}$ hold?
May
28
asked $\sigma$-algebra and measurable set.
May
27
awarded  Popular Question
May
26
comment Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
@ReneSchipperus because $\varphi(x)=\frac{1}{x}$. then $F(x,\varphi(x))=\frac{1}{\sqrt{(1-\frac{1}{x^2})(1-1)}}$ so it is meaningless
May
26
comment Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
@DavidH I have edited the post. the integral lower bound is $1$, and $x \in (0,1)$. then the integral is real-valued.
May
26
comment Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
@JPi, sorry .I made a mistake. the integral is from $1$,not $0$
May
26
revised Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
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