Aditya

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visits	member for	2 years, 9 months
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	169 reputation	bio	website	randomdeterminism.wordpress.c…	visits	member for	2 years, 9 months
5 badges		location			seen	Apr 23 at 23:40

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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 $[0,1]$
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.
Mar 27	awarded	Supporter