# Probability & Statistics - Assignment

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Statistics and Probability
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Question1a)The algebraic formula s for S is S = i=1NYib)The contracts are likely to be executed per dayE(s) = E[E(s)]=E¿¿E()¿μE(N)c)Daily executed contractsVar(s) = E¿= E¿Here we have used the fact that Yi’s are independent = E(Nμ2)+μ2var(N)= E(Nσ2)+μ2var(N)= σ2E(N)+μ2var(N)Therefore, Var(s)=σ2E(N)+μ2var(N)d)Probability distribution If the probability of Yi = yi would be defined as P(Yi = yi) = pi. The probability would satify the following;0<pi<1 for each ip1+ p2+ ... + pi= 1.e) Mean = μEach value of Yi’s would be multiplied by its probabilityP* Yi’sThe value would then be summed upPYiThe standard deviation = σEach value of Yi’s would be squared and multiplied by its probabilityP* Yi2The value would then be summed upPYi2The square of expected value = μ2would then be subtracted from the summed up value
PYi2¿μ2The standard deviation would be determined by finding its square root σ=PYi2μ2Question 21.Unconditional mean and variancePt = pt-1 + tPt-1 = pt-2 + t-1Pt = p0 + (t + t-1 + t-2 ...,)E(Pt) = E(p0) + E(t + t-1 + t-2 ...,c )E(P0) = 0E(Pt) = t*μE(¿=μV(Pt) = V(p0) + V(t + t-1 + t-2 ...,c )Since t are unconditional μV(t + t-1) = V(t) + V(t-1)V(Pt) = t * σ22.Condition mean and variance Pt = Pt-1 + tPt-1 = xE(Pt) = E(Pt-1) + E(Et) μE(Pt) = x + μV(Pt) = V(Pt-1) + V(Pt) V(Pt-1) = 0V(Pt) = σ23.Variance in ‘1’ depends on t and is sum of variance for ‘t’ periods while in one it is equal variance of ‘1’ period Question 3a)¿3)3 - 3 ́r¿=¿3 - ́r¿ + ¿2 - ́r¿+¿1 - ́r¿ = ¿3 _ r2 + r1 - 3 ́r¿2
= r32 + r2r3 + r1r3 - 3 ́rr3 + r2r3 + r22 + r2r1 + 3 ́rr2 +r3r1 = r2r1 + r12 + 3 ́rr1 - 3 ́rr3 - 3 ́rr2 - 3 ́rr1 + 9 ́r2= r3(r3 + 2r2) - 3 ́r¿r3 - 3 ́r¿ + r2(r2 + 2r1)b)t=1T¿¿¿¿ = r3(r3+2r2)3 ́r(2r33 ́r)+r2(r2+2r1)T(3) +r3(r3+2r2)3 ́r(2r33 ́r)+r2(r2+2r1)T(2) + r3(r3+2r2)3 ́r(2r33 ́r)+r2(r2+2r1)T(1)c)If statistically S2(3) and ́r are independent when r1, r2 ...., rnN(μ,σ2)d)S2(1) = t=1T¿¿¿¿ S2(3) = t=1T¿¿¿¿ Cross multiplying = s2(3)¿¿¿¿S2(3) = 3*t=1T¿¿¿¿

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