Question 1: Dependence
* Let X be a Binomial random variable defined as the sum of 6 independent Bernoulli trials. The probability of a Bernoulli taking the value 1 is given by p. Suppose that prior to the 6 Bernoulli trials, p is chosen to take one of three possible values with the following probabilities:
. . .p = 0.2 with probability 0.1
. . .p = 0.6 with probability 0.7
. . .p = 0.8 with probability 0.2
* Compute the joint probability distribution of X and p. Are X and p independent? Prove your reasoning.
* Compute the unconditional mean and variance of X.
* Compute the conditional mean of X given each possible value of p. Based on your calculations, what sign do you expect the covariance between X and p to be?
Above is the given question..
I'm finding it difficult to find X and p as there are 3 different p's to find..
Here's my working out so far but not sure if I'm on the right track:
n=6
x=1
p=.2 or .6 or .8
P(.2)=.3932160 x .1 =.0393216
P(.6)=.0368640 x .7 =.0258048
P(.8)=.0015360 x .2 =.0003072
X=6 x P(.2) = .2359296
X=6 x P(.6) = .1548288
X=6 x P(.8) = .0018432
I don't really know where to go from here, dont know how I can form a joint probability table as I have multiple values... any help or suggestions would be greatly appreciated! Thanks
* Let X be a Binomial random variable defined as the sum of 6 independent Bernoulli trials. The probability of a Bernoulli taking the value 1 is given by p. Suppose that prior to the 6 Bernoulli trials, p is chosen to take one of three possible values with the following probabilities:
. . .p = 0.2 with probability 0.1
. . .p = 0.6 with probability 0.7
. . .p = 0.8 with probability 0.2
* Compute the joint probability distribution of X and p. Are X and p independent? Prove your reasoning.
* Compute the unconditional mean and variance of X.
* Compute the conditional mean of X given each possible value of p. Based on your calculations, what sign do you expect the covariance between X and p to be?
Above is the given question..
I'm finding it difficult to find X and p as there are 3 different p's to find..
Here's my working out so far but not sure if I'm on the right track:
n=6
x=1
p=.2 or .6 or .8
P(.2)=.3932160 x .1 =.0393216
P(.6)=.0368640 x .7 =.0258048
P(.8)=.0015360 x .2 =.0003072
X=6 x P(.2) = .2359296
X=6 x P(.6) = .1548288
X=6 x P(.8) = .0018432
I don't really know where to go from here, dont know how I can form a joint probability table as I have multiple values... any help or suggestions would be greatly appreciated! Thanks
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