# Probability

Problem 1

Suppose that X, Y, and Z are independent, with E[X]=E[Y]=E[Z]=2, and E[X2]=E[Y2]=E[Z2]=5.

Find cov(XY,XZ).

cov(XY,XZ)= ?

Problem 2.

Let X be a standard normal random variable. Another random variable is determined as follows. We flip a fair coin (independent from X). In case of Heads, we let Y=X. In case of Tails, we let Y=−X.

1) Is Y normal? Justify your answer.
yes
no
not enough information to determine

2) Compute Cov(X,Y).
Cov(X,Y)= ?

Are X and Y independent?
yes
no
not enough information to determine

Problem 3.

2.0 points possible (graded, results hidden)
Find P(X+Y≤0).

P(X+Y≤0)= ?

1. 👍 1
2. 👎 0
3. 👁 580
1. Use Cov(XY,XZ)=E( XY XZ)E(XY)E(XZ) =E(X2)E(Y)E(Z)E(X)E(Y)E(X)E(Z). =5
Is that helpful?

1. 👍 3
2. 👎 9
2. you post a complicated statistical formula, yet you have trouble with a simple quadratic equation?

What's wrong with this picture?

1. 👍 2
2. 👎 2
3. E[XYXZ]-E[XY]*E[XZ]= E[X^2]*E[Y]*E[Z]-E[X]*E[Y]*E[X]*E[Z]

Because X is independent of Y and Z, X^2 is also independent of Y and Z

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2. 👎 0
4. is the Y is normal

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2. 👎 0
5. Y is normal, did anyone found the Cov(X,Y)? Or the probability?

I found that Cov=1 and P(X+Y)=3/4

But I am not sure...

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2. 👎 0
6. Can you write steps to your solution?

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2. 👎 0
7. X is a standard normal, E[X]=0 and Var(X)=1

Cov(X,Y) = E[XY] - E[X]E[Y]. Since Y=|X|, Cov(X,Y) = E[X^2] - E[X].E[X] = Var(X) --> E[X^2] = 1 --> Cov(X,Y) = 1. They are not independent because is different than 0.

The probability question I found it intuitively,
for all values of X=-Y, the results are 0, so 1/2(fair coin toss)*1

for X=Y, half of the values are minor or equal to 0, so 1/2(fair coin toss)*1/2

P(X+Y)= (1/2*1) + (1/2*1/2) = 3/4

Does that makes sense?

1. 👍 1
2. 👎 0
8. I think that it does.

1. 👍 0
2. 👎 0

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