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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.
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
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
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
comment Mutually exclusive events
$E^c = \{3,4,5,6\} \not\equiv \{3,4\} = F$
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
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 $n$.
Apr
26
comment Given P(A|B) and P(A|C), how to get or strategically approach P(A|(B & C))?
Try expanding P(A,B,C) using the law of total probability in different ways.
Aug
8
comment A probability game
Sorry. I didn't realize the difference.