Mixing
Joint Distribution of two dependent Bernoulli Random Variables for $rho=1$
$begingroup$
Say we have two random variables $Xsim B(p_1), Ysim B(p_2)$ where $B(p)=$ Bernoulli with probability $0le ple 1$. I am interested in the case when the correlation $rho$ of $X,Y$ tends to $1$.
If we set events $A={X=1}$, $B={Y=1}$, the conditional probability property gives us
$$begin{align}tag{1}
P(Acap B) = P(Amid B)cdot P(B)\ = P(Bmid A)cdot P(A)
end{align}$$
When $rho=1$ we must have $P(Amid B)=P(Bmid A)=1$ (since one event implies the other). Equation $(1)$ then yields
$$P(A) = P(B)$$
Furthermore, if correlation is high (= tending to 1), then whenever event $A$ occurs then $B$ must also occur (and vice versa). So the probabilities of $A,B$ should be
$$P(A)=P(B)=max(p_1,p_2)tag{2}$$
However, for any $rho=1-epsilon$ with $epsilon>0$, we still have $P(A)=p_1$ and $P(B)=p_2$ as per definition.
So can $(2)$ even be correct? What am I missing?
Unfortunately, Joint distribution of dependent Bernoulli Random variables only discusses non-deterministic sequences, so it doesn't quite apply.
probability-distributions random-variables correlation
asked Jan 25 at 13:30
Phil-ZXXPhil-ZXX
1,33611538
$endgroup$
add a comment |
$begingroup$
Say we have two random variables $Xsim B(p_1), Ysim B(p_2)$ where $B(p)=$ Bernoulli with probability $0le ple 1$. I am interested in the case when the correlation $rho$ of $X,Y$ tends to $1$.
If we set events $A={X=1}$, $B={Y=1}$, the conditional probability property gives us
$$begin{align}tag{1}
P(Acap B) = P(Amid B)cdot P(B)\ = P(Bmid A)cdot P(A)
end{align}$$
When $rho=1$ we must have $P(Amid B)=P(Bmid A)=1$ (since one event implies the other). Equation $(1)$ then yields
$$P(A) = P(B)$$
Furthermore, if correlation is high (= tending to 1), then whenever event $A$ occurs then $B$ must also occur (and vice versa). So the probabilities of $A,B$ should be
$$P(A)=P(B)=max(p_1,p_2)tag{2}$$
However, for any $rho=1-epsilon$ with $epsilon>0$, we still have $P(A)=p_1$ and $P(B)=p_2$ as per definition.
So can $(2)$ even be correct? What am I missing?
Unfortunately, Joint distribution of dependent Bernoulli Random variables only discusses non-deterministic sequences, so it doesn't quite apply.
probability-distributions random-variables correlation
asked Jan 25 at 13:30
Phil-ZXXPhil-ZXX
1,33611538
$endgroup$
$begingroup$
"we still have $P(A)=p_1$ and $P(B)=p_2$". Try $p_1=0.2$ and $p_2=0.8$ and try to get a value of $rho$ being 0.9. Can't do it. Given $p_1$ and $p_2$, there are certain values of $rho$ that can't be obtained. So you can't fix $p_1$ and $p_2$ and take the limit of $rho$ -> $1$. Hint: start with the definition of $rho$.
$endgroup$
– JimB
Jan 25 at 14:36
add a comment |
$begingroup$
Say we have two random variables $Xsim B(p_1), Ysim B(p_2)$ where $B(p)=$ Bernoulli with probability $0le ple 1$. I am interested in the case when the correlation $rho$ of $X,Y$ tends to $1$.
If we set events $A={X=1}$, $B={Y=1}$, the conditional probability property gives us
$$begin{align}tag{1}
P(Acap B) = P(Amid B)cdot P(B)\ = P(Bmid A)cdot P(A)
end{align}$$
When $rho=1$ we must have $P(Amid B)=P(Bmid A)=1$ (since one event implies the other). Equation $(1)$ then yields
$$P(A) = P(B)$$
Furthermore, if correlation is high (= tending to 1), then whenever event $A$ occurs then $B$ must also occur (and vice versa). So the probabilities of $A,B$ should be
$$P(A)=P(B)=max(p_1,p_2)tag{2}$$
However, for any $rho=1-epsilon$ with $epsilon>0$, we still have $P(A)=p_1$ and $P(B)=p_2$ as per definition.
So can $(2)$ even be correct? What am I missing?
Unfortunately, Joint distribution of dependent Bernoulli Random variables only discusses non-deterministic sequences, so it doesn't quite apply.
probability-distributions random-variables correlation
asked Jan 25 at 13:30
Phil-ZXXPhil-ZXX
1,33611538
$endgroup$
Say we have two random variables $Xsim B(p_1), Ysim B(p_2)$ where $B(p)=$ Bernoulli with probability $0le ple 1$. I am interested in the case when the correlation $rho$ of $X,Y$ tends to $1$.
If we set events $A={X=1}$, $B={Y=1}$, the conditional probability property gives us
$$begin{align}tag{1}
P(Acap B) = P(Amid B)cdot P(B)\ = P(Bmid A)cdot P(A)
end{align}$$
When $rho=1$ we must have $P(Amid B)=P(Bmid A)=1$ (since one event implies the other). Equation $(1)$ then yields
$$P(A) = P(B)$$
Furthermore, if correlation is high (= tending to 1), then whenever event $A$ occurs then $B$ must also occur (and vice versa). So the probabilities of $A,B$ should be
$$P(A)=P(B)=max(p_1,p_2)tag{2}$$
However, for any $rho=1-epsilon$ with $epsilon>0$, we still have $P(A)=p_1$ and $P(B)=p_2$ as per definition.
So can $(2)$ even be correct? What am I missing?
Unfortunately, Joint distribution of dependent Bernoulli Random variables only discusses non-deterministic sequences, so it doesn't quite apply.
probability-distributions random-variables correlation
probability-distributions random-variables correlation
asked Jan 25 at 13:30
Phil-ZXXPhil-ZXX
1,33611538
asked Jan 25 at 13:30
Phil-ZXXPhil-ZXX
1,33611538
asked Jan 25 at 13:30
Phil-ZXXPhil-ZXX
1,33611538
asked Jan 25 at 13:30
Phil-ZXXPhil-ZXX
1,33611538
asked Jan 25 at 13:30
Phil-ZXXPhil-ZXX
1,33611538
1,33611538
$begingroup$
"we still have $P(A)=p_1$ and $P(B)=p_2$". Try $p_1=0.2$ and $p_2=0.8$ and try to get a value of $rho$ being 0.9. Can't do it. Given $p_1$ and $p_2$, there are certain values of $rho$ that can't be obtained. So you can't fix $p_1$ and $p_2$ and take the limit of $rho$ -> $1$. Hint: start with the definition of $rho$.
$endgroup$
– JimB
Jan 25 at 14:36
add a comment |
$begingroup$
"we still have $P(A)=p_1$ and $P(B)=p_2$". Try $p_1=0.2$ and $p_2=0.8$ and try to get a value of $rho$ being 0.9. Can't do it. Given $p_1$ and $p_2$, there are certain values of $rho$ that can't be obtained. So you can't fix $p_1$ and $p_2$ and take the limit of $rho$ -> $1$. Hint: start with the definition of $rho$.
$endgroup$
– JimB
Jan 25 at 14:36
$begingroup$
"we still have $P(A)=p_1$ and $P(B)=p_2$". Try $p_1=0.2$ and $p_2=0.8$ and try to get a value of $rho$ being 0.9. Can't do it. Given $p_1$ and $p_2$, there are certain values of $rho$ that can't be obtained. So you can't fix $p_1$ and $p_2$ and take the limit of $rho$ -> $1$. Hint: start with the definition of $rho$.
$endgroup$
– JimB
Jan 25 at 14:36
$begingroup$
"we still have $P(A)=p_1$ and $P(B)=p_2$". Try $p_1=0.2$ and $p_2=0.8$ and try to get a value of $rho$ being 0.9. Can't do it. Given $p_1$ and $p_2$, there are certain values of $rho$ that can't be obtained. So you can't fix $p_1$ and $p_2$ and take the limit of $rho$ -> $1$. Hint: start with the definition of $rho$.
$endgroup$
– JimB
Jan 25 at 14:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The apparent inconsistency is due to the fact that $rho$ cannot take on all values between $-1$ and $1$ given specific values for $p_1$ and $p_2$.
Using the definition of a correlation coefficient
$$rho={{Cov(x_1,x_2)} over {sqrt{Var(x_1)Var(x_2)}}}={{text{Pr}(x_1=1,x_2=1)-p_1 p_2}over{sqrt{p_1(1-p_1)p_2(1-p_2)}}}$$
and noting that $text{Pr}(x_1=1,x_2=1)leq text{Min}(p_1,p_2)$ we can find the maximum value of $rho$ for all possible values of $p_1$ and $p_2$ (here using Mathematica):
sol = Maximize[{(P11 - p1 p2)/Sqrt[p1 (1 - p1) p2 (1 - p2)],
0 < p1 < 1 && 0 < p2 < 1 && 0 < P11 <= Min[p1, p2]}, P11][[1]]
$$begin{array}{cc}
{ &
begin{array}{cc}
sqrt{frac{text{p1} (text{p2}-1)}{(text{p1}-1) text{p2}}} & left(0<text{p2}<frac{1}{2}land 0<text{p1}leq text{p2}right)lor left(frac{1}{2}leq text{p2}<1land text{p1}=text{p2}right)lor left(frac{1}{2}leq text{p2}<1land 0<text{p1}<text{p2}right) \
sqrt{frac{(text{p1}-1) text{p2}}{text{p1} (text{p2}-1)}} & left(0<text{p2}<frac{1}{2}land text{p2}<text{p1}<1right)lor left(frac{1}{2}leq text{p2}<1land text{p2}<text{p1}<1right) \
-infty & text{True} \
end{array}
\
end{array}$$
ContourPlot[sol, {p1, 0, 1}, {p2, 0, 1}, ContourLabels -> True]
In short, $rho=1$ can only occur when $p_1=p_2$.
answered Jan 25 at 17:30
JimBJimB
58037
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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votes
$begingroup$
The apparent inconsistency is due to the fact that $rho$ cannot take on all values between $-1$ and $1$ given specific values for $p_1$ and $p_2$.
Using the definition of a correlation coefficient
$$rho={{Cov(x_1,x_2)} over {sqrt{Var(x_1)Var(x_2)}}}={{text{Pr}(x_1=1,x_2=1)-p_1 p_2}over{sqrt{p_1(1-p_1)p_2(1-p_2)}}}$$
and noting that $text{Pr}(x_1=1,x_2=1)leq text{Min}(p_1,p_2)$ we can find the maximum value of $rho$ for all possible values of $p_1$ and $p_2$ (here using Mathematica):
sol = Maximize[{(P11 - p1 p2)/Sqrt[p1 (1 - p1) p2 (1 - p2)],
0 < p1 < 1 && 0 < p2 < 1 && 0 < P11 <= Min[p1, p2]}, P11][[1]]
$$begin{array}{cc}
{ &
begin{array}{cc}
sqrt{frac{text{p1} (text{p2}-1)}{(text{p1}-1) text{p2}}} & left(0<text{p2}<frac{1}{2}land 0<text{p1}leq text{p2}right)lor left(frac{1}{2}leq text{p2}<1land text{p1}=text{p2}right)lor left(frac{1}{2}leq text{p2}<1land 0<text{p1}<text{p2}right) \
sqrt{frac{(text{p1}-1) text{p2}}{text{p1} (text{p2}-1)}} & left(0<text{p2}<frac{1}{2}land text{p2}<text{p1}<1right)lor left(frac{1}{2}leq text{p2}<1land text{p2}<text{p1}<1right) \
-infty & text{True} \
end{array}
\
end{array}$$
ContourPlot[sol, {p1, 0, 1}, {p2, 0, 1}, ContourLabels -> True]
In short, $rho=1$ can only occur when $p_1=p_2$.
answered Jan 25 at 17:30
JimBJimB
58037
$endgroup$
add a comment |
$begingroup$
The apparent inconsistency is due to the fact that $rho$ cannot take on all values between $-1$ and $1$ given specific values for $p_1$ and $p_2$.
Using the definition of a correlation coefficient
$$rho={{Cov(x_1,x_2)} over {sqrt{Var(x_1)Var(x_2)}}}={{text{Pr}(x_1=1,x_2=1)-p_1 p_2}over{sqrt{p_1(1-p_1)p_2(1-p_2)}}}$$
and noting that $text{Pr}(x_1=1,x_2=1)leq text{Min}(p_1,p_2)$ we can find the maximum value of $rho$ for all possible values of $p_1$ and $p_2$ (here using Mathematica):
sol = Maximize[{(P11 - p1 p2)/Sqrt[p1 (1 - p1) p2 (1 - p2)],
0 < p1 < 1 && 0 < p2 < 1 && 0 < P11 <= Min[p1, p2]}, P11][[1]]
$$begin{array}{cc}
{ &
begin{array}{cc}
sqrt{frac{text{p1} (text{p2}-1)}{(text{p1}-1) text{p2}}} & left(0<text{p2}<frac{1}{2}land 0<text{p1}leq text{p2}right)lor left(frac{1}{2}leq text{p2}<1land text{p1}=text{p2}right)lor left(frac{1}{2}leq text{p2}<1land 0<text{p1}<text{p2}right) \
sqrt{frac{(text{p1}-1) text{p2}}{text{p1} (text{p2}-1)}} & left(0<text{p2}<frac{1}{2}land text{p2}<text{p1}<1right)lor left(frac{1}{2}leq text{p2}<1land text{p2}<text{p1}<1right) \
-infty & text{True} \
end{array}
\
end{array}$$
ContourPlot[sol, {p1, 0, 1}, {p2, 0, 1}, ContourLabels -> True]
In short, $rho=1$ can only occur when $p_1=p_2$.
answered Jan 25 at 17:30
JimBJimB
58037
$endgroup$
add a comment |
$begingroup$
The apparent inconsistency is due to the fact that $rho$ cannot take on all values between $-1$ and $1$ given specific values for $p_1$ and $p_2$.
Using the definition of a correlation coefficient
$$rho={{Cov(x_1,x_2)} over {sqrt{Var(x_1)Var(x_2)}}}={{text{Pr}(x_1=1,x_2=1)-p_1 p_2}over{sqrt{p_1(1-p_1)p_2(1-p_2)}}}$$
and noting that $text{Pr}(x_1=1,x_2=1)leq text{Min}(p_1,p_2)$ we can find the maximum value of $rho$ for all possible values of $p_1$ and $p_2$ (here using Mathematica):
sol = Maximize[{(P11 - p1 p2)/Sqrt[p1 (1 - p1) p2 (1 - p2)],
0 < p1 < 1 && 0 < p2 < 1 && 0 < P11 <= Min[p1, p2]}, P11][[1]]
$$begin{array}{cc}
{ &
begin{array}{cc}
sqrt{frac{text{p1} (text{p2}-1)}{(text{p1}-1) text{p2}}} & left(0<text{p2}<frac{1}{2}land 0<text{p1}leq text{p2}right)lor left(frac{1}{2}leq text{p2}<1land text{p1}=text{p2}right)lor left(frac{1}{2}leq text{p2}<1land 0<text{p1}<text{p2}right) \
sqrt{frac{(text{p1}-1) text{p2}}{text{p1} (text{p2}-1)}} & left(0<text{p2}<frac{1}{2}land text{p2}<text{p1}<1right)lor left(frac{1}{2}leq text{p2}<1land text{p2}<text{p1}<1right) \
-infty & text{True} \
end{array}
\
end{array}$$
ContourPlot[sol, {p1, 0, 1}, {p2, 0, 1}, ContourLabels -> True]
In short, $rho=1$ can only occur when $p_1=p_2$.
answered Jan 25 at 17:30
JimBJimB
58037
$endgroup$
The apparent inconsistency is due to the fact that $rho$ cannot take on all values between $-1$ and $1$ given specific values for $p_1$ and $p_2$.
Using the definition of a correlation coefficient
$$rho={{Cov(x_1,x_2)} over {sqrt{Var(x_1)Var(x_2)}}}={{text{Pr}(x_1=1,x_2=1)-p_1 p_2}over{sqrt{p_1(1-p_1)p_2(1-p_2)}}}$$
and noting that $text{Pr}(x_1=1,x_2=1)leq text{Min}(p_1,p_2)$ we can find the maximum value of $rho$ for all possible values of $p_1$ and $p_2$ (here using Mathematica):
sol = Maximize[{(P11 - p1 p2)/Sqrt[p1 (1 - p1) p2 (1 - p2)],
0 < p1 < 1 && 0 < p2 < 1 && 0 < P11 <= Min[p1, p2]}, P11][[1]]
$$begin{array}{cc}
{ &
begin{array}{cc}
sqrt{frac{text{p1} (text{p2}-1)}{(text{p1}-1) text{p2}}} & left(0<text{p2}<frac{1}{2}land 0<text{p1}leq text{p2}right)lor left(frac{1}{2}leq text{p2}<1land text{p1}=text{p2}right)lor left(frac{1}{2}leq text{p2}<1land 0<text{p1}<text{p2}right) \
sqrt{frac{(text{p1}-1) text{p2}}{text{p1} (text{p2}-1)}} & left(0<text{p2}<frac{1}{2}land text{p2}<text{p1}<1right)lor left(frac{1}{2}leq text{p2}<1land text{p2}<text{p1}<1right) \
-infty & text{True} \
end{array}
\
end{array}$$
ContourPlot[sol, {p1, 0, 1}, {p2, 0, 1}, ContourLabels -> True]
In short, $rho=1$ can only occur when $p_1=p_2$.
answered Jan 25 at 17:30
JimBJimB
58037
edited Jan 26 at 1:34
answered Jan 25 at 17:30
JimBJimB
58037
answered Jan 25 at 17:30
JimBJimB
58037
answered Jan 25 at 17:30
JimBJimB
58037
58037
add a comment |
add a comment |
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$begingroup$
"we still have $P(A)=p_1$ and $P(B)=p_2$". Try $p_1=0.2$ and $p_2=0.8$ and try to get a value of $rho$ being 0.9. Can't do it. Given $p_1$ and $p_2$, there are certain values of $rho$ that can't be obtained. So you can't fix $p_1$ and $p_2$ and take the limit of $rho$ -> $1$. Hint: start with the definition of $rho$.
$endgroup$
– JimB
Jan 25 at 14:36