For binomial distribution, E(X) is the expectation. How to induce E(X^2) in terms of E(X). Thanks
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For binomial distribution, E(X) is the expectation. How to induce E(X^2) in terms of E(X). Thanks
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- Suppose that the random change in value of a financial asset is X over the first day and Y over the second. Suppose also that Var(X) =18 and Var(Y) = 26 In this case, the total change in the value over these two days is given by X +Y. Do you have enough information to compute Var(X +Y)? If so, compute this value. If not, explain what additional information you need to do so.Suppose that f (x) = e^-x for 0<x. Determine the following probabilities:b) Explain how X can be related to the exponential distribution by using a moment generating function argument.
- Find the probability that the first traffic light is green given thatthe second traffic light is green.I Know: P(A) = 0.55. P(A^c) = 0.45 P(B|A) = 0.75 P(B|A^c) = 0.09 P(B) = 0.453I Want to find: P(A|B)For a continuous pdf, the probability of an exact value can be found . Is this true or false ? And example ?g(X)=x^2+1 A=5 X ~ exp(1) E [g (X) | {X> 3}] conditional expected Calculate the value.
- Use the geometric distribution to derive E(Y), E(Y^2), and V(Y) from the Poisson distributionThe number of work accidents that occur per week in a mine follows Poisson's law so that the probability of 2 accidents occurring is equal to 3/2 of the probability of an accident occurring. Calculate the probability that the time for the first accident to occur is one week.Calculate the mean and variance when the probability variable X's moment-generating function is as follows f(x)= 2x-1/16 , x=1,2,3,4 my answer is mean = 25/8, var = 170/16 - (50/16)^2 but solution is mean = 2, var = 4/5 How do we solve the problem? Help me