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seen Aug 27 at 4:45

Jun
6
comment Is there a measure on $\mathbb{N}$? (What is the chance that a random integer from $\mathbb{N}$ is even?)
@EricTowers, the counting measure is not a probability measure.
Jun
6
asked Counterexample for converse about measurable sections
Jul
7
comment solving a problem in probability
Nitpicking: weight cannot be negative, hence the Gaussian assumption is unrealistic.
Jun
23
comment Mutually exclusive events
Then you should assume $P(E) = P(F) = 0.5$
Jun
22
comment Mutually exclusive events
The event that $E$ does not occur first is (in my notaton) $A^c$. You cannot simply change the meaning of $E$ (which is an event in experiment $\mathcal E_1$). Perhaps the solution given by @DilipSarwate is close to what you are thinking: Think of the experiment in which either $E$ or $F$ occur for the first time. What is the probability that the event that occurred was $E$.
Jun
22
comment What is the best choice in a win/lose game and how to calculate it
Agreed. To explain buying of lottery, you need to add additional utility for participating in the game---the thrill of the possibility of winning.
Jun
22
awarded  Commentator
Jun
22
comment Mutually exclusive events
Does my updated answer clarify this point?
Jun
22
comment What is the best choice in a win/lose game and how to calculate it
(Tongue in cheek comment): Math doesn't help because math assumes linear utilities. Assume concave utilities, and math will explain risk adverse behavior and buying of lotteries.
Jun
22
awarded  Editor
Jun
22
comment Mutually exclusive events
If $P(E) = P(F) = 1$, then $E$ and $F$ cannot be mutually exclusive because $E \cup F \subset \Omega$, thus $P(E \cup F) = P(E) + P(F) \le P(\Omega) = 1$.
Jun
22
revised Mutually exclusive events
More explanation
Jun
22
comment Mutually exclusive events
$E^c = \{3,4,5,6\} \not\equiv \{3,4\} = F$
Jun
22
answered Mutually exclusive events
May
17
comment Lower bound on a minimum of maximum of a sequence of standard normal random variables
In the RHS of all equations, $j$ should be replaced by $p$.
Apr
26
comment Show that if $\{x_{n}\}$ and $\{y_{n}\}$ are Cauchy sequences in X then $d(x_{n},y_{n})$ converges in $\mathbb{R}$.
In the reverse triangle inequality, the RHS should be $d(x,z)$.
Oct
23
comment Roots of $f_n(x) = 1 + (1-x)^2 - (x+3)(1-x)^{n+1}$ in the interval $[0,1]$
Yes, it is easy to see that the roots form a decreasing sequence. $f_{n+1}(x) - f_n(x) > 0$. If $a_n$ is a root of $f_n$ then, $f_{n+1}(a_n) > 0$. Thus, there is a root of $f_{n+1}$ between $0$ (where $f_{n+1}$ is negative) and $a_n$ (where $f_{n+1}$ is positive).
Oct
23
awarded  Student
Oct
23
comment Roots of $f_n(x) = 1 + (1-x)^2 - (x+3)(1-x)^{n+1}$ in the interval $[0,1]$
@SL2:The derivative is not always positive. The derivative is $[ (n+2)x + 3n + 2 ](1-x)^n - 2(1-x)$ and its sign is determined by whether $a_n(x) = [(n+2)x + 3n + 2 ](1-x)^{n-1}$ is greater than or less than $2$. As $n \to \infty$, $a_n(x) \to 0$, for $x \in [0,1]$. Thus, the derivative, $a_n(x) - 2$, is negative for large $n$.
Oct
23
asked Roots of $f_n(x) = 1 + (1-x)^2 - (x+3)(1-x)^{n+1}$ in the interval $[0,1]$