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43m
answered Find $ 1 - a^x$, where $x$ is a random variable
5h
answered Strenuous, arduous, laborious, onerous sum
6h
answered Proving $P(A|(B \cap C)) = P(B | (A \cap C)) P(A | C) / P(B | C)$ using Bayes' theorem.
6h
comment Is $f(x)=f(x+2\pi)$?
Yes: if $f(x)$ is only defined on $(-\pi,0)\cup(0,\pi)$ then $f(x)-f(x+2\pi)$ is not meaningful for any value of $x$
19h
comment Why does the cross product follow the right hand rule?
That is due to a rotation invariance
19h
answered Why does the cross product follow the right hand rule?
21h
comment Does $1-1+1/2-1/2+1/3-1/3+\cdots$ converges?
The sum of the first one term is $1$ which is $\dfrac{2}{1+1}$. The sum of the first three terms is $\dfrac12$ which is $\dfrac{2}{3+1}$. The sum of the first five terms is $\dfrac13$ which is $\dfrac{2}{5+1}$ and similarly for other odd numbers
22h
comment Does $1-1+1/2-1/2+1/3-1/3+\cdots$ converges?
+1: in a similar vein, if there are $n$ terms then the partial sum is $\frac{2}{n+1}$ or $0$ depending on whether $n$ is odd or even, i.e. $\frac{1 -(-1)^n}{n+1}$.
22h
answered Statistics: Simple pick from bag problem
2d
answered finding and angle and coordinate point
2d
comment Find $E[Z_1 | aZ_1 + bZ_2]$
It could be anything. In particular, given any non-zero $a, b, E[Z_1], E[Z_2], Var(Z_1), Var(Z_2), s$ and target $t$ it would be possible to find distributions of $Z_1$ and $Z_2$ with $E[Z_1 | aZ_1 + bZ_2=s]=t$
2d
revised Combinations from prime number of elements
deleted 76 characters in body
2d
answered Combinations from prime number of elements
2d
comment Conditional probability problem and Alias Method
I thought the "alias method" was for generating random variates. Do you plan to simulate this? Calculation would be easier and more accurate.
2d
comment Conditional probability problem and Alias Method
I do not think it is clear, but I would read it as asking for the conditional probability that at least three calls arrive given that at least two calls arrive.
2d
comment What does “combining the solutions in O(n) time” mean?
I think you may need to to put $O(\cdots )$ on the right hand side, or at least multiply $n \log n$ and $n^2$ by unknown constants
2d
comment Probability distribtuions
The exact distribution is binomial, so you are looking for an approximation for that. There are at least two possibilities, one closer than the other.
2d
comment Number of possibilities to draw from a card deck isn't an integer - where's my error?
Combinations or binomial coefficients
2d
comment Number of possibilities to draw from a card deck isn't an integer - where's my error?
So the same answer as Brian M. Scott, but with a different approach
2d
answered Number of possibilities to draw from a card deck isn't an integer - where's my error?