If $(X_t)_tI$ Is a Martingale and $$ Is a Stopping Time with $T$, then $X_text EX_Tmidmathcal F_$. W

Equation $(4)$ is not true if $Ica(-infty,T$ is uncountale. Secifically, your first equality is actually using countale additivity of measures, which of course requires the index set to e countale. It doesn't really have much to do with $taut$ having measure zero, although it is not difficult to show that $t:mathrm Ptaut0$ is at most countale for any random variale $tau$.

Related Questions

more than usual vs. more than usuallyThe question is whether you mean to say that the numer of customers is more than the usual numer (adjective) or that you had more customers than you usually have (adver). So either is correct. Practically seaking, they amount to the same thing. We (in AmE, and er David Pugh, in BrE as well) tend to use the shorter form "usual", and not dwell on this sutle distinction.

Accurately recalling the key - can everyone do it?

Even studies on the Levitin Effect, a phenomenon that posits that people can tend to accurately recall the key of a familiar melody, discover that at least a significant minority of people cannot produce this effect (see here for an example), and this effect has been found to be hard to reproduce. So, I'd say that it's actually not that common for people to accurately recall the key

Printing to a Cups Server without using the local CUPS doesn't work since OS X Lion

Set up /etc/cups/client.conf correctly for your environment so that the cups printers display in the print config dialog.To get the queues to display in the applications as a choice, make each printer the default printer once. Distribute the resulting com.apple.print.favorites.plist to users with mcx or copy to /Library/Preferences It uses the PPDs on the server. No client config necessary!.

How to find the intersection part in a Venn diagram?

You have a big set of Students $S$ and you have two subsets $A,B$ of students who study chemistry or physics respectively. You know that $#(Acup B)20-317$.

Also note that $Acap B (A^ccup B^c)^c$ where $c$ denotes the complement. But you do know how many people do NOT study chemistry or physics.Do you know how to continue from here?

Suppose that G is a group and N Try to show that there is a correspondence between subgroups of $G/N$ and subgroups of $G$ containing $N$. Also show that this correspondence preserves notions of normality.In fact, you have hinted at this correspondence in your question: a subgroup $ g_1N, dots , g_k N$ of $G/N$ corresponds to the subgroup $g_1N cup dots cup g_k N$ of $G$, which clearly contains $N$.

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Are there other simple conditions we can use to demonstrate the non-existence of a universal set?

We can use Cantor's Theorem actually.Cantor's Theorem states that for any set $X$, cardinality of power set $|P(X)| > |X|$.Now, suppose for a contradiction that set of all sets $S$ exists. But then $P(S) subseteq S$ because every set in $P(S)$ also included in $S$ by definition of $S$. But then $|P(S)| le |S|$ and by Cantor's Theorem $|P(S)| > |S|$, a contradiction

How to Start Program After SSHD is Up and Running

The init.d scripts combined with update-rc.d were unsuccessful in rendering my desired system behavior. I ended up learning more about UpStart, which seems to be the primary method for starting programs on Ubuntu.I created this script in /etc/init as start_program.

conf:I then just needed to run the command:And after a reboot everything worked as I had desired in my initial post

Proving that a punctured disk is not simply connected, using a specific definition

Note that the punctured disk is punctured, in that it does not contain $0$. Further, you can enclose $0$ in a small open ball $B$, and $ z:|z| geq 1 $ in the complement of a closed ball. Calling this ball $C$, we need it to be such that $B cap C^complement emptyset$. The complement of $H$, which is $ 0 cup z: |z| geqslant 1 $, is contained in these two disjoint open sets

Calculate P(A|AUBUC) independent events

Alternatively:

$$P(Acup Bcup C)P(A)P(B)P(C)-P(Acap B)-P(Acap C)-P(Bcap C)P(Acap Bcap C)0.70.40.3-0.28-0.21-0.120.0840.874.$$

Hence:

$$P(A|Acup Bcup C)fracP(Acap (Acup Bcup C))P(Acup Bcup C)fracP(A)P(Acup Bcup C)frac0.70.874approx0.8.$$

Cannot connect cups client with server

I'm not sure if this was your problem but this may help someone:If you're getting the lpstat: Error - add '/version1.1' to the server name error when you've added '/version1.1to the server URI in yourlpstatcommand or in/etc/cups/cupsd.

conf, make sure the server is up, running, and accessible! It threw the error for me until I realized the server was down

Cursor movement related tput commands under zsh: can the 'clear' behavior be configured?

zsh outputs its completions below the prompt, so it makes sure that area is clear.I don't think you can disable it. However you can tell zsh that the escape sequence to clear until the end of the screen is the empty string.Then, you can start zsh with:And unset TERMINFO later, but you'll find that the completion is never cleared which makes it awkward to use.

Proving $(Atimes C)cup(Btimes D)subseteq(Acup B)times(Ccup D)$

Suppose $(x,y)in (Atimes C)cup (Btimes D)$. Then $xin A$ or $xin B$, and $yin C$ or $yin D$. Indeed, then $xin Acup B, yin Ccup D$. So, $(x,y)in (Acup B)times (Ccup D)$. This implies that

$$ (Atimes C)cup (Btimes D)subseteq (Acup B)times (Ccup D). $$

Quotient map $q:X to X/A$ is open if $A$ is open (?)

That argument is correct, and generalises to the case of finitely many closed sets being identified to a point (more common than "pressed") and then the quotient map is closed.Identifying open sets to a point, as in $Bbb R/(0,1)$ is uncommon because we lose properties like $T_1$, so the resulting quotient is already non-metrisable etc. Closed sets (or even compact ones) is much more common in practice.

Counting some polynomials that have a zero in $mathbbZ_nX$

I think there's a simple answer when $n$ is prime. Count instead the polynomials that don't have a zero. Such a polynomial must map $lbrace0,1,dots,n-1rbrace$ to $lbrace1,dots,n-1rbrace$. There are $(n-1)^n$ such maps. But each of those maps corresponds to a unique polynomial, since Lagrange interpolation works over a field. So the number you are looking for is $n^n-(n-1)^n$

If you learn that the individual did not like vehicle $#1$, what now is the probability that he/she liked at least one of the other two vehicles?

I am stuck on the fourth part. What I have done so far : I am thinking P(A2 | A1') P(A3 | A1') P(A3 intersect A2 | A1')Close. You want: $quadmathsf P(A_2cup A_3mid A_1') mathsf P(A_2mid A_1') mathsf P(A_3mid A_1') - mathsf P(A_2cap A_3 mid A_1')$ Can you find these terms ?

Why defining regular conditional probability?

Conditional probabilites and conditional expectations are only defined up to null sets. In many problems you would like to put many null sets involved in conditional expectations inside one null set, but uncountable unions of null sets need not be null. When we are dealing with Borel measures on 'nice spaces' (say complete separable metric spaces) it is possible to handle these null sets using regular conditional probabilities.

How can I explain the difference between NULL and zero?

Really it isn't relevant whether your boss is a programmer. The issue is a conceptual one not a technical one.Ask him to assume you got a raise. Your old salary was 175k, but your new salary is unknown. Then ask him - what percentage raise did you receive?If he is arithmetically disabled, walk him through the process until he can see where the missing link is

Show that if $A$ and $B$ are sets, then $(Acap B) cup (Acap overlineB)A$.

Following your approach of chasing elements, you can say If $x in (A cap overlineB)$ it means $x in A wedge x in overlineB$, so $x in A wedge x not in B$. Therefore $x in ((Acap B) cup (Acap overlineB))$ means $(x in A wedge x in B)vee (x in A wedge x not in B)$ and use the distributive principle

Explain in simple terms : $P to (Q lor R) $ is equivalent to $P landlnot Q to R$

Look at these two truth tables: $$beginarray|c|c|c|c|c| hline % after : hline or clinecol1-col2 clinecol3-col4 ... P & Q & R & Q cup R :I & PLongrightarrow I hline 0 & 0 & 1 & 1 & 1 0 & 0 & 1 & 1 & 1 0 & 1 & 0 & 1 & 1 0 & 1 & 1 & 1 & 1 1 & 0 & 0 & 0 & 0 1 & 0 & 1 & 1 & 1 1 & 1 & 0 & 1 & 1 1 & 1 & 1 & 1 & 1 hline

endarray$$$$beginarray|c|c|c|c|c| hline % after : hline or clinecol1-col2 clinecol3-col4 ... P & Q' & R & P cap Q' :J & JLongrightarrow R hline 0 & 1 & 1 & 0 & 1 0 & 1 & 1 & 0 & 1 0 & 0 & 0 & 0 & 1 0 & 0 & 1 & 0 & 1 1 & 1 & 0 & 1 & 0 1 & 1 & 1 & 1 & 1 1 & 0 & 0 & 0 & 1 1 & 0 & 1 & 0 & 1 hline

endarray$$

Use of Sure in reply to help offering and to appreciation

"Sure" in the first usage is an emphatic "yes"; it implies that the second speaker is sure they will like a cup of water."Sure" in the second usage is a shortening of "sure thing", which in turn is a shortening of "it was a sure thing", implying that no thanks are necessary because the second speaker was certain to perform the action for which they're being thanked

Are these disjoint/dependent?

You are correct and so are your derivations, except that $P(Acap B).4$ not 0.55.$P(A^ccap B^c).11-P(Acup B)1-P(A)P(B)-P(Acap B) rightarrow 0.9.7.6-P(Acup B) rightarrow P(A cup B) .7.6-.9 .4 neq .7*.6 .42$Disjoint events have additive probabilities for their union and depnendent events are simply not independent, so you verified this witha a counterexample.

Comma before a participial phrase

the real problem with "I was given a cup made in China from my friend" is that it is in the passive voice. If I were editing a paper, I would recommend changing it to the active voice. My friend gave me a cup made in China; though my personal preference would be to include "which.

" So, "My friend gave me a cup which was made in China."

If $int_E fint_E g$ then $fg$ a.e.?

Why do you claim a contradiction in the last two lines? The hypothesis is that

$$int_E f int_E g$$

but it need not be the case that

$$int_E_n f int_E_n g$$

Consider, for example, $E 0,2pi$, $f(x) cos(x)$, and $g(x) sin(x)$. Then $int_E f int_E g 0$ but, for example, $int_0^pi f neq int_0^pi g$.

probability question involving set theory

Write $A^ccup B$ as a disjoint union:$$A^ccup B(A^csetminus B)cup(Bsetminus A^c)cup (A^ccap B)(A^ccap B^c)cup (Bcap A)cup (A^ccap B).$$Since $Asubseteq B$, so $A^ccup BOmega$ and therefore $$OmegaB^ccup Acup (A^ccap B)$$ from which the result follows

find a condition on B,C so set difference is associative (AB)CA(BC)

Another way:Since $(Asetminus B)setminus C A setminus (Bcup C)$, the associativity equation can equivalently be written as

$$Asetminus (Bcup C) Asetminus (Bsetminus C)$$

Now obviously that is true for all $A$ exactly if

$$Bcup C Bsetminus C$$

But those two sets differ exactly by the elements of $C$, thus they are equal if and only if $Cemptyset$.

Advice for making/keeping shredded chicken moist?

Chicken breast is not suitable for making shredded meat. For that, you need collagen-rich dark meat, for example chicken thighs. If you cook your chicken breast less, as moscafj suggested, you can certainly get tasty chicken breast. For easy portioning, you can precut it into strips and keep a supply of pan-fried strips, for example. But you will never get it to the point where it shreds properly

Two events A,B. Given $P(B)$, $P(A|B)$, and $P(B^c|A^c)$, need to find P(B|A)

Continuing from where you left off we have $$P(B^c|A^c)frac 1-P(A)-P(B)P(A cap B) 1-P(A).$$ Multiply both sides by $1-P(A)$ an write the equation as $$P(A)1-P(B^c|A^c)1-P(B)P(A cap B)-P(B^c|A^c).$$ Now divide by $1-P(B^c|A^c)$ to get the value of $P(A)$.

Prove the following set identity using the laws of set theory

Suppose there exists $x in$ the set. Then$x in (B cup C)'$$x in A cup (B cup C)$ but $x notin (B cup C)$ so $x in A$$x in A' cup (B cup C)$ but $x notin (B cup C)$ so $x in A'$So $x in A cap A' emptyset$.A contradiction. So the set has no elements. So the set is empty.