Probability / Statistics 2?

Can't figure these out, please help.





4. Suppose X and Y be independent normal random variables, with distributions N(μ, sigm) = N(5, 3) and N(2, 4) respectively. Let W = X + 2Y + 1.


Compute


(a) E(W); (b) V (W); (c) Prob(W %26lt; 0) (d) Cov(X,W) .


5. Suppose that X and Y are rv’s which only assume the values 0 and 1. Suppose that the joint pmf p(x, y) satisfies p(0, 0) = .3, p(0, 1) = .4, p(1, 0) = .2.


(a) What is p(1, 1)?


(b) Compute E(2X − Y ).


(c) Are X and Y independent?


(d) Compute the conditional pmf pX(x).


6. Suppose that the number of phone calls Tara receives in the evening is a Poisson process with rate 2 per hour, and the number of calls I receive is an independent Poisson process with rate 0.1 per hour.


(a) At 6 p.m., Tara returns home and happily waits for her next


phone call. What is the probability that she will get a call before 7:30 pm?


(b) What is the probability that at least one of us receives at least one phone call between 6 and 7:30 pm?

Probability / Statistics 2?
While your question is mostly clear I'm just going to clarify that I read the distributions as mean and variance, so X has a mean of 5 and a variance of 3.





4.


Remember that expectation is a linear operator.





E(W) = E(X+2Y+1) = E(X) + 2E(Y) + 1 = 5 + 2*2 + 1 = 10





Because the variables are independent you the variances add.


Additive constants have no effect on the variance as variance is a measure of spread and shifting the data dose not affect the spread. Multiplicative constants come out squared.





Var(W) = Var(X+2Y+1) = Var(X+2Y) = Var(X) + 4*Var(Y) = 3 + 4*4 = 19





W~N(10,19)





P(W%26lt;0) = P((W-10)/Sqrt(19) %26lt; -10/Sqrt(19)) = P(Z%26lt;-2.29) = 0.011 where Z is the standard normal.





because X and Y are independent the covariance is 0.





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5





only options are p(0,0), p(1,0), p(0,1) and p(1,1). the total probability of all four event has to be equal to one, thus p(1,1) = 0.1





P(X = 0) = 0.7


P(X = 1) = 0.3


P(Y = 0) = P(Y=1) = 0.5





E(X) = 0.3


E(Y) = 0.5


E(2X-Y) = 0.1





P(X=0|Y=0) = .3/.5 = .6 %26lt;%26gt; P(X=0) so no, the variables are not independent. (%26lt;%26gt; means not equal to)





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6.





T~Poisson(2t), Y~Poisson(0.1t)





a) T~Poisson(2*1.5)=Poisson(3)


P(T = 1) = 3e^-3=0.149


P(T %26gt;0 ) = 1-e^-3 = 0.95





b) T+Y~Poisson(3.15)


P(T+Y %26gt; 0) = 1-e^(-3.15) = 0.957
Reply:Too much work for 10 points.