What is the Premack Differential probability principle?
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In operant conditioning, the Premack principle, developed by David Premack states that a commonly occurring action (one more desirable for the actor) can be used effectively as a reinforcer for a less commonly occurring one (that is, one less desirable for the actor). A common example used to illustrate this principle is a parent requiring a child to clean his or her room before he or she can watch television. In this case, television, an activity that probably does not require reinforcement, is used as a reinforcer for cleaning the room, which in the context of this example the child would not do without reinforcement.\n\nA reinforcement hierarchy can be used to determine the relative frequency and desirability of different actions, and is employed when applying the Premack principle in operant conditioning. This would be a list of actions, starting with the most desirable and ending with the least desirable. In the example, watching television would be close to the top of the list whereas cleaning the room would be closer to the bottom.
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evenly distributed energy
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(A) The probability that a wait will be necessary is the same as the probability that a car comes within the next T seconds: 1 - exp(-lambda T). So the probability of no wait is:\n\nP = exp(-lambda*T)\n\n(B) Let Q be the expected waiting time. If t is the time until the next car arrives (t < T), the expected additional wait time will be Q, giving an expected total wait time of t + Q. In other words the expectation value of the total wait time is:\n\nQ = &#92;integral_0^T (lambda exp(-lambda*t))(t + Q) dt\n= [-exp(-lambda t)(Q + 1/lambda + t]_0^T\n= Q + 1/lambda - exp(-lambda T)(Q + 1/lambda + T)\n\nSolving for Q gives the answer:\n\nQ = (exp(+lambda T) - 1)/lambda - T
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This is true of any distribution that has a density, not just the normal one. It follows straight from the definitions. The c.d.f. (cumulative distribution function) of a vble X is by def. the fctn F(a) = Prob(X < a). Density, g, is the derivative of F: g(a) is by def. the limit of (1/epsilon)Prob(a < X < a+epsilon) as epsilon decreases to zero. So F(a) is the integral of g from -infty to a (this uses F(-infty) = 0 and the Fundamental Thm of Calculus, that integration reverses differentiation).
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This is an easy derivation. \n\nMultiplicative Principle: \nOne fundamental principle of counting is that if you have n of one type of object (N) and m of another type of object (M), then the number of ways to pair objects of type one with objects of type two with regard to order is n*m. This is argued by saying that one can first pair one of the N-type objects with all of the M type objects for m pairs, and then do the same for the remaining N type objects, giving m + m + ... + m n times, or n*m pairings. \n\nPermutations:\nSuppose you have k objects and you want to know how many ways you can order them in a line. You have k ways to choose the first object, k - 1 ways of choosing the second object, and so forth until you run out of objects. By the fundamental principles of counting, this makes the number of possible orderings k*(k - 1)*...*1 = k! which we will call P(k).\n\nCombinations:\nSuppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted (by the fundamental principles of counting). Let P(k) be the number of ways of ordering k objects (this is just k! = k*(k-1)*...*1 ). Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!].
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let us denote the marginal density of X_i by f_i, that is the density of X_1 is f_1.\nlet us denote the density of the triplet (X_1, X_2 X_3) by f.\n\nas X_1, X_2, X_3 are stochasticaly independent,\n\nf(x_1, x_2, x_3) = f_1(x_1)  * f_2(x_2) *  f3(x_3)\n\nlet g be a bounded borelian function on R ^3\n\nset:\nphi(x_1, x_2, x_3) = (x_1 + x_2 + x_3) ^ 2 / 6\n\nE(g o phi (X_1, X_2, X_3)) = &#92;int g o phi(x_1, x_2, x_3) f(x_1, x_2, x_3) dx_1 dx_2 dx_3\n\n = &#92;int g( ( x_1 + x_2 + x_3) ^ 2 / 6) f(x_1, x_2, x_3) dx_1dx_2dx_3\n\nthen you may conclude using Fubini theorem
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M(t), the moment gerenating function associated to a pdf (probability distribution funciton) is given by\nm(t)=E(e^(tX))= Integral( e^(tx) f(x) dx) from -inf to + inf.\nif it exists, them m(t) is continuously diffrentiable in a neighborhood of the origin, and \nthe nth derivative of m(t), evaluated at t=0 is equal to the nth moment; that is, to E(X^n).
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try\nhttp://www.peterwebb.co.uk/probability.htm
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Well,\n           natural BCS are Neumann BCs; i.e. fixing the gradient of the unknown at the boundaries.\n\n          In contract, essential BCS are Dirichlet BCs i.e. fixing the solution at the boundaries\n\nBye\nRavi
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Random chances are the same as probabilities. If you are so interested, you should look into Quantum Theory. It describes it in depth.
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Look up convergence theory on www.colorado.edu/communication/met.  It will explain this theory in fairly simple terms.
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This is an inevitable occurrence; there is actually a theorem by Paul Erdos (I forget who he was working with) that is applicable...  Erdos-Schekeres Theorem, perhaps.  It's the one where they proved that if you take a piece of paper, cover the bottom of a box with it, remove it, wad it up and throw it back into the box, at least one point will be exactly above the same place it was before...  It doesn't take a whole lot of imagination to see a definite isomorphism between the E-S theorem (or whichever one it is) and what you're talking about.\n\nANYHOW, what it boils down to, in your case, is that you will have an experience (whether in dream or in reality is not germain) and it will match a future one.  On the whole, you should expect that to happen once in a while--and if you have a wide enough breadth of experience, perhaps more often...\n\nI'm not sure that this addresses your question.  If it doesn't, here's a shorter answer: It is part of normal life; it isn't a problem; and we all share it to some degree or other.
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I could tell you the answer or I could give you an idea of what to consider when you do it yourself.\n\na) Binomial numbers and p = 0.7, q= 0.3 might help here.  You do not know the size of the population of the mailing list so you might need to make an assumption that it is large enough to assume a Normal distribution.  I think the central limit theorem might help here.\n\nb) Binomial theory, and conditional probabilities.  Look it up in your text book!  Again if the size of the population is large enough you might be able to simplify the problem.
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x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
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Oy that is a great question.  I take it you taking a risk managment or a derivative course! ;)  Basically you would have to put in functions (functions of interest and volatility variations) for both and then evaluate it accordingly.  It is nearly impossible to do by hand but computer can provide you with some good insight :).  That's when 3D probability that engineers love so much kicks in.  \n\nRival
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Great question.  Mathematically, it is expedient to calculate probabilities with a infinite pool of possible results.  However, causality rules the day for organizing those possibilities into realistic probabilities.  So, one can safely say that the probability of all the QED probabilities inherent in your hand and the wall suddenly violating the molecular and chemical structures involved would be equally infinite.  It is unreasonable to estimate.  Just like the ridiculous hundred monkeys producing Shakespeare if given enough time.  The reality is this: a hundred monkeys in one room die of pretty quickly.  Oh, what about just a random stream of letters?  Who constructs this random string?  And keeps it running?  They will all be extinct before it spits out half a page of text.  Two probabilities that need infinite time to realize a result do not sum to a possibility in this case.
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Probability function of a poission distribution\n==> P(X=x) = e^(-m)*m^(x)/x! \n[x= 0,1,2,.... ]  countably infinite\n\nMean = 4 ==> m = 4\n\nProbability that x = 2\n==> P(X=2)= e^(-4)*4^(2)/2!\n          = \n\nProbability that x <= 2\n==> P(X<=2)= P(X=0) + P(X=1) + P(X=2)\n==> P(X<=2)\n= e^(-4)*4^(0)/0! + e^(-4)*4^(1)/1! + e^(-4)*4^(2)/2!\n= e^(-4) {1/1 + 4/1 + 16/2}\n= e^(-4) {1 + 4 + 8}\n= e^(-4) {13}\n\nProbability that x > 2\n==> P(X > 2)\n= P(X=3) + P(X=4) + ...... \n(Or) = 1 - {P(X<=2)}\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/