Show that $X$, $Y$, and $Z$ are pairwise independent. Are they independent?
$begingroup$
Let $X$ and $Y$ be independent random variables, each taking the values $-1$ or $1$ with probability $1/2$, and let $Z = XY$. Show that $X$, $Y$, and $Z$ are pairwise independent. Are they independent?
MY ATTEMPT
To begin with, let us determine the distribution of $Z$: if $X = -1$ and $Y = -1$ or $X = 1$ and $Y = 1$, then $Z = 1$. If $X = -1$ and $Y = 1$ or $X = 1$ and $Y = -1$, then $Z = -1$. Since we considered all possible cases, here is the distribution of $Z$:
begin{align*}
p_{Z}(1) = textbf{P}(Z = 1) = textbf{P}(X = -1)textbf{P}(Y = -1) + textbf{P}(X = 1)textbf{P}(Y = 1) = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
p_{Z}(-1) = textbf{P}(Z = -1) = textbf{P}(X = -1)textbf{P}(Y = 1) + textbf{P}(X = 1)textbf{P}(Y = -1) = frac{1}{2}
end{align*}
Therefore $Zsim Bernoulli(1/2)$.
As to the second part of the question, we have
begin{align*}
p_{X,Z}(1,1) & = textbf{P}(Z = 1 mid X = 1)textbf{P}(X = 1) = textbf{P}(XY = 1mid X = 1)textbf{P}(X = 1)\
& = textbf{P}(Y = 1mid X = 1)textbf{P}(X = 1) = textbf{P}(Y = 1)textbf{P}(X = 1) = frac{1}{4} = p_{X}(1)p_{Z}(1)
end{align*}
Analogous reasoning applies to the remaining cases. However, the random variables $X$, $Y$ and $Z$ are not independent, because
begin{align*}
p_{X,Y,Z}(1,-1,1) = textbf{P}(Z = 1mid{X = 1}cap{Y = -1})textbf{P}(Y = -1mid X = 1)textbf{P}(X = 1) = 0
end{align*}
My question is: are my results correct? Can it be improved? Any contribution is appreciated. Thanks in advance.
probability probability-theory proof-verification random-variables
asked Jan 23 at 15:48
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be independent random variables, each taking the values $-1$ or $1$ with probability $1/2$, and let $Z = XY$. Show that $X$, $Y$, and $Z$ are pairwise independent. Are they independent?
MY ATTEMPT
To begin with, let us determine the distribution of $Z$: if $X = -1$ and $Y = -1$ or $X = 1$ and $Y = 1$, then $Z = 1$. If $X = -1$ and $Y = 1$ or $X = 1$ and $Y = -1$, then $Z = -1$. Since we considered all possible cases, here is the distribution of $Z$:
begin{align*}
p_{Z}(1) = textbf{P}(Z = 1) = textbf{P}(X = -1)textbf{P}(Y = -1) + textbf{P}(X = 1)textbf{P}(Y = 1) = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
p_{Z}(-1) = textbf{P}(Z = -1) = textbf{P}(X = -1)textbf{P}(Y = 1) + textbf{P}(X = 1)textbf{P}(Y = -1) = frac{1}{2}
end{align*}
Therefore $Zsim Bernoulli(1/2)$.
As to the second part of the question, we have
begin{align*}
p_{X,Z}(1,1) & = textbf{P}(Z = 1 mid X = 1)textbf{P}(X = 1) = textbf{P}(XY = 1mid X = 1)textbf{P}(X = 1)\
& = textbf{P}(Y = 1mid X = 1)textbf{P}(X = 1) = textbf{P}(Y = 1)textbf{P}(X = 1) = frac{1}{4} = p_{X}(1)p_{Z}(1)
end{align*}
Analogous reasoning applies to the remaining cases. However, the random variables $X$, $Y$ and $Z$ are not independent, because
begin{align*}
p_{X,Y,Z}(1,-1,1) = textbf{P}(Z = 1mid{X = 1}cap{Y = -1})textbf{P}(Y = -1mid X = 1)textbf{P}(X = 1) = 0
end{align*}
My question is: are my results correct? Can it be improved? Any contribution is appreciated. Thanks in advance.
probability probability-theory proof-verification random-variables
asked Jan 23 at 15:48
user1337user1337
46110
$endgroup$
1
$begingroup$
Looks correct okay to me.
$endgroup$
– drhab
Jan 23 at 16:02
add a comment |
$begingroup$
Let $X$ and $Y$ be independent random variables, each taking the values $-1$ or $1$ with probability $1/2$, and let $Z = XY$. Show that $X$, $Y$, and $Z$ are pairwise independent. Are they independent?
MY ATTEMPT
To begin with, let us determine the distribution of $Z$: if $X = -1$ and $Y = -1$ or $X = 1$ and $Y = 1$, then $Z = 1$. If $X = -1$ and $Y = 1$ or $X = 1$ and $Y = -1$, then $Z = -1$. Since we considered all possible cases, here is the distribution of $Z$:
begin{align*}
p_{Z}(1) = textbf{P}(Z = 1) = textbf{P}(X = -1)textbf{P}(Y = -1) + textbf{P}(X = 1)textbf{P}(Y = 1) = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
p_{Z}(-1) = textbf{P}(Z = -1) = textbf{P}(X = -1)textbf{P}(Y = 1) + textbf{P}(X = 1)textbf{P}(Y = -1) = frac{1}{2}
end{align*}
Therefore $Zsim Bernoulli(1/2)$.
As to the second part of the question, we have
begin{align*}
p_{X,Z}(1,1) & = textbf{P}(Z = 1 mid X = 1)textbf{P}(X = 1) = textbf{P}(XY = 1mid X = 1)textbf{P}(X = 1)\
& = textbf{P}(Y = 1mid X = 1)textbf{P}(X = 1) = textbf{P}(Y = 1)textbf{P}(X = 1) = frac{1}{4} = p_{X}(1)p_{Z}(1)
end{align*}
Analogous reasoning applies to the remaining cases. However, the random variables $X$, $Y$ and $Z$ are not independent, because
begin{align*}
p_{X,Y,Z}(1,-1,1) = textbf{P}(Z = 1mid{X = 1}cap{Y = -1})textbf{P}(Y = -1mid X = 1)textbf{P}(X = 1) = 0
end{align*}
My question is: are my results correct? Can it be improved? Any contribution is appreciated. Thanks in advance.
probability probability-theory proof-verification random-variables
asked Jan 23 at 15:48
user1337user1337
46110
$endgroup$
Let $X$ and $Y$ be independent random variables, each taking the values $-1$ or $1$ with probability $1/2$, and let $Z = XY$. Show that $X$, $Y$, and $Z$ are pairwise independent. Are they independent?
MY ATTEMPT
To begin with, let us determine the distribution of $Z$: if $X = -1$ and $Y = -1$ or $X = 1$ and $Y = 1$, then $Z = 1$. If $X = -1$ and $Y = 1$ or $X = 1$ and $Y = -1$, then $Z = -1$. Since we considered all possible cases, here is the distribution of $Z$:
begin{align*}
p_{Z}(1) = textbf{P}(Z = 1) = textbf{P}(X = -1)textbf{P}(Y = -1) + textbf{P}(X = 1)textbf{P}(Y = 1) = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
p_{Z}(-1) = textbf{P}(Z = -1) = textbf{P}(X = -1)textbf{P}(Y = 1) + textbf{P}(X = 1)textbf{P}(Y = -1) = frac{1}{2}
end{align*}
Therefore $Zsim Bernoulli(1/2)$.
As to the second part of the question, we have
begin{align*}
p_{X,Z}(1,1) & = textbf{P}(Z = 1 mid X = 1)textbf{P}(X = 1) = textbf{P}(XY = 1mid X = 1)textbf{P}(X = 1)\
& = textbf{P}(Y = 1mid X = 1)textbf{P}(X = 1) = textbf{P}(Y = 1)textbf{P}(X = 1) = frac{1}{4} = p_{X}(1)p_{Z}(1)
end{align*}
Analogous reasoning applies to the remaining cases. However, the random variables $X$, $Y$ and $Z$ are not independent, because
begin{align*}
p_{X,Y,Z}(1,-1,1) = textbf{P}(Z = 1mid{X = 1}cap{Y = -1})textbf{P}(Y = -1mid X = 1)textbf{P}(X = 1) = 0
end{align*}
My question is: are my results correct? Can it be improved? Any contribution is appreciated. Thanks in advance.
probability probability-theory proof-verification random-variables
probability probability-theory proof-verification random-variables
asked Jan 23 at 15:48
user1337user1337
46110
asked Jan 23 at 15:48
user1337user1337
46110
asked Jan 23 at 15:48
user1337user1337
46110
asked Jan 23 at 15:48
user1337user1337
46110
asked Jan 23 at 15:48
user1337user1337
46110
46110
1
$begingroup$
Looks correct okay to me.
$endgroup$
– drhab
Jan 23 at 16:02
add a comment |
1
$begingroup$
Looks correct okay to me.
$endgroup$
– drhab
Jan 23 at 16:02
1
1
$begingroup$
Looks correct okay to me.
$endgroup$
– drhab
Jan 23 at 16:02
$begingroup$
Looks correct okay to me.
$endgroup$
– drhab
Jan 23 at 16:02
add a comment |
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1
$begingroup$
Looks correct okay to me.
$endgroup$
– drhab
Jan 23 at 16:02