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Jul
1
awarded  Enlightened
Jun
30
awarded  Nice Answer
Jun
9
reviewed Close compute standard basis in local rings
Jun
9
reviewed Close In a 30-60 right triangle the side opposite the 30 degree angle is half the length of the hypotenuse. Why?
Jun
9
reviewed Close Is there a formal definition for antiderivatives?
Jun
9
reviewed Close Convergence of sequence of function in norm.
Jun
8
comment Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$
No, $n \approx 1.7305073578576$.
Jun
5
comment Inside a circle, radius 1, 10 points are chosen. $X$- random variable that represents the distance from the rim of the circle to the closest point.
Yours is the same as mine. In your case $n=10$ and the event $\{A\}$ is exactly $D > d$.
Jun
5
comment Maximum of a multivariable function
The answer to this problem (the maximum of $f$ subject to this condition) can be expressed in algebraic numbers only (as a root of a polynomial of order 8). Are you sure there is no typo? Try Maximize[{y^2 - 2 z^2 + 2 x^2 y, 0 <= x <= 1 && 0 <= y <= 1 && 0 <= z <= 1 && x == z + y Sqrt[1 - y^2]}, {x, y, z}, Reals] // FullSimplify if you have access to Mathematica.
Jun
5
answered Inside a circle, radius 1, 10 points are chosen. $X$- random variable that represents the distance from the rim of the circle to the closest point.
Jun
5
revised The probability of more than $10$ people getting their own hat is no more than $1/100$
added 4 characters in body
Jun
2
revised Expectation of the min of two independent random variables?
fixed typo in the text
Jun
2
comment PDF of a sum of exponential random variables
It is called the law of total expectation. Indeed $f_Y(y) = \Pr(Y=y) = \mathbb{E}\left([Y=y]\right) = \mathbb{E}\left(\mathbb{E}\left([Y=y] \mid N\right) \right) = \mathbb{E}\left(f_{Y|N}\left(y\right) \right)$.
May
30
revised Infinite sum of reciprocals of pentagonal numbers
deleted 2 characters in body
May
30
answered Infinite sum of reciprocals of pentagonal numbers
May
29
comment Joint probability distribution (over unit circle)
By using $\{\sqrt{x^2+y^2} \leqslant u\} \equiv \{ -u \leqslant x \leqslant u, -\sqrt{u^2 - x^2} \leqslant y \leqslant \sqrt{u^2-x^2} \}$.
May
29
answered Joint probability distribution (over unit circle)
May
29
revised Joint probability distribution (over unit circle)
added 14 characters in body; edited title
May
20
revised bound on expectation of a two-variable function under an independent distribution
added 12 characters in body
May
18
awarded  Yearling