Can you please help answer
B) Compute Cov(X,Y)= ?
are X and Y independent?
a) yes
b) no
c) not enough information
C) Find P(X+Y<=0) =?
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
Is Y normal? (1 option is correct).
a) yes
b) no
c) not enough information
Compute Cov(X,Y)= ?
are X and Y independent?
a) yes
b) no
c) not enough information
5 answers
not so sure, what do you think about this reasoning:
B) Cov(X,Y)= ?
if Heads with p=1/2 (fair coin): Cov(X,Y) = E[XY] - E[X]E[Y] = E[XY] = E[X^2] = var(X) -(E[X])^2 = var(X) = 1
if Tails with p=1/2: Cov(X,Y) = E[-X^2] = - E[X^2] = -var(X) = - 1
Cov(X,Y)= 1/2*1 + 1/2*-1 = 0
X and Y are independent if P(Y │X) = P (Y). if we know x, we know y is either + or - x, so X gives us some useful information about Y. Therefore X and Y are not independent?
C) Find P(X+Y<=0) = ?
if Heads with p=1/2: Y=X => P(X+Y<=0) = P(X+X<=0) = P(2X<=0) = P(X<=0) =1/2 (cause standard normal r.v.)
if Tails with p=1/2: Y=-X => P(X+Y<=0) = P(X-X<=0) = P(0<=0) = 1
P(X+Y<=0) = 1/2*1/2 + 1/2*1 = 3/4
B) Cov(X,Y)= ?
if Heads with p=1/2 (fair coin): Cov(X,Y) = E[XY] - E[X]E[Y] = E[XY] = E[X^2] = var(X) -(E[X])^2 = var(X) = 1
if Tails with p=1/2: Cov(X,Y) = E[-X^2] = - E[X^2] = -var(X) = - 1
Cov(X,Y)= 1/2*1 + 1/2*-1 = 0
X and Y are independent if P(Y │X) = P (Y). if we know x, we know y is either + or - x, so X gives us some useful information about Y. Therefore X and Y are not independent?
C) Find P(X+Y<=0) = ?
if Heads with p=1/2: Y=X => P(X+Y<=0) = P(X+X<=0) = P(2X<=0) = P(X<=0) =1/2 (cause standard normal r.v.)
if Tails with p=1/2: Y=-X => P(X+Y<=0) = P(X-X<=0) = P(0<=0) = 1
P(X+Y<=0) = 1/2*1/2 + 1/2*1 = 3/4
I would argue something different.
Since X is standard normal r.v. which means that it is symmetric around 0, then it has the following PDF:
fx(x) = 1/sqrt(2pi) * exp{-x^2/2}.
Now, if our coin toss happen to be heads (with probab. 1/2) then our PDF of Y is exactly the same as above. What happens however if we get tails?
Since, X is symmetric then Y=-X is exactly the same as Y=X, which can be shown as follows:
fy(-x) = 1/sqrt(2pi) * exp {-(-x)^2/2} = 1/sqrt(2pi) * exp {-x^2/2) = fy(x)=fx(x).
The above then implies that Y=X no matter the results of the coin toss, so Y is normal and the cov(X,Y) = cov(X,X) = var(X) = 1, as X is standard normal. Finally, since X and Y are identical, then both X and Y are dependent.
In the end P(X+Y<=0) is P(2X<=0), which is the same as P(X<=0), so it is equal to 0.5.
Since X is standard normal r.v. which means that it is symmetric around 0, then it has the following PDF:
fx(x) = 1/sqrt(2pi) * exp{-x^2/2}.
Now, if our coin toss happen to be heads (with probab. 1/2) then our PDF of Y is exactly the same as above. What happens however if we get tails?
Since, X is symmetric then Y=-X is exactly the same as Y=X, which can be shown as follows:
fy(-x) = 1/sqrt(2pi) * exp {-(-x)^2/2} = 1/sqrt(2pi) * exp {-x^2/2) = fy(x)=fx(x).
The above then implies that Y=X no matter the results of the coin toss, so Y is normal and the cov(X,Y) = cov(X,X) = var(X) = 1, as X is standard normal. Finally, since X and Y are identical, then both X and Y are dependent.
In the end P(X+Y<=0) is P(2X<=0), which is the same as P(X<=0), so it is equal to 0.5.
I do not agree with the answer above.
In both cases (H or T) Y is distributed with the same pdf, but the value of Y is different (-X or X). So we cannot say that two rv.s are the same if they have the same pdf.
Y and X are of course dependent. If we draw from the unconditional probability P(Y) we can get whatever Y. If we draw form P(Y|X=x) we are sure that we get either X=x or X=-x.
In both cases (H or T) Y is distributed with the same pdf, but the value of Y is different (-X or X). So we cannot say that two rv.s are the same if they have the same pdf.
Y and X are of course dependent. If we draw from the unconditional probability P(Y) we can get whatever Y. If we draw form P(Y|X=x) we are sure that we get either X=x or X=-x.
Correct ans:
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 .
Is 𝑌 normal?
yes
Compute 𝖢𝗈𝗏(𝑋,𝑌) .
𝖢𝗈𝗏(𝑋,𝑌)= 0
Are 𝑋 and 𝑌 independent?
no
Find 𝑃(𝑋+𝑌≤0).
𝑃(𝑋+𝑌≤0)= 3/4
First, observe that 𝑋+𝑌 has a symmetric distribution, that is,
𝑃(𝑋+𝑌≤𝑐)=𝑃(𝑋+𝑌≥−𝑐),
for any 𝑐 . This is because with probability 1/2 , 𝑋+𝑌=0 , and with probability 1/2 , 𝑋+𝑌 is a normal of variance 4 .
Thus,
𝑃(𝑋+𝑌=0) =1/2,
𝑃(𝑋+𝑌≠0) =1/2,
𝑃(𝑋+𝑌<0) =𝑃(𝑋+𝑌>0)=1/4.
This gives 𝑃(𝑋+𝑌≤0)=𝑃(𝑋+𝑌<0)+𝑃(𝑋+𝑌=0)=3/4 .
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 .
Is 𝑌 normal?
yes
Compute 𝖢𝗈𝗏(𝑋,𝑌) .
𝖢𝗈𝗏(𝑋,𝑌)= 0
Are 𝑋 and 𝑌 independent?
no
Find 𝑃(𝑋+𝑌≤0).
𝑃(𝑋+𝑌≤0)= 3/4
First, observe that 𝑋+𝑌 has a symmetric distribution, that is,
𝑃(𝑋+𝑌≤𝑐)=𝑃(𝑋+𝑌≥−𝑐),
for any 𝑐 . This is because with probability 1/2 , 𝑋+𝑌=0 , and with probability 1/2 , 𝑋+𝑌 is a normal of variance 4 .
Thus,
𝑃(𝑋+𝑌=0) =1/2,
𝑃(𝑋+𝑌≠0) =1/2,
𝑃(𝑋+𝑌<0) =𝑃(𝑋+𝑌>0)=1/4.
This gives 𝑃(𝑋+𝑌≤0)=𝑃(𝑋+𝑌<0)+𝑃(𝑋+𝑌=0)=3/4 .
8 answers