X has a Bernoulli distribution with parameter $p$ and Y has a Poisson distribution with mean $p$. What is the...
$begingroup$
X ~ Bernoulli($p$) and Y ~ Poisson($p$) where $0<p<1$ and $p$ is the fixed number.
What is the maximum value of $P(X=Y)$ where the maximum value is taken over all possible joint distributions with specified marginal distributions?
That is... what is the joint distribution that maximizes the probability of X being equal to Y?
If I assume that $X$ and $Y$ are independent, finding the probability is easy since the only possible value $X$ can be is either $0$ or $1$. However, if I don't assume their independence, things get complicated. Any hints or help please?
probability probability-theory probability-distributions
edited Jan 13 at 20:43
Bernard
119k740113
asked Jan 13 at 20:24
JHKJHK
12
$endgroup$
add a comment |
$begingroup$
X ~ Bernoulli($p$) and Y ~ Poisson($p$) where $0<p<1$ and $p$ is the fixed number.
What is the maximum value of $P(X=Y)$ where the maximum value is taken over all possible joint distributions with specified marginal distributions?
That is... what is the joint distribution that maximizes the probability of X being equal to Y?
If I assume that $X$ and $Y$ are independent, finding the probability is easy since the only possible value $X$ can be is either $0$ or $1$. However, if I don't assume their independence, things get complicated. Any hints or help please?
probability probability-theory probability-distributions
edited Jan 13 at 20:43
Bernard
119k740113
asked Jan 13 at 20:24
JHKJHK
12
$endgroup$
$begingroup$
Since $P(X=0)=1-p<e^{-p}=P(Y=0)$, $P(X=1)=p>pe^{-p}=P(Y=1)$, and $P(Xgeqslant2)=0$, the maximum is $$1-p+pe^{-p}$$
$endgroup$
– Did
Jan 13 at 20:30
$begingroup$
@Did That comment should be an answer, I think
$endgroup$
– leonbloy
Jan 13 at 20:31
$begingroup$
@Did Are we adding the lower probability for each case to ensure the equality? like.. adding the upper bound for each case?
$endgroup$
– JHK
Jan 13 at 20:45
$begingroup$
Use the remark that $$P(X=Y)=sum_nP(X=n,Y=n)leqslantsum_nmin{P(X=n),P(Y=n)}$$ hence in the present case, $$P(X=Y)leqslant P(X=0)+P(Y=1)$$ and to complete the proof, it remains to build $(X,Y)$ such that the inequality is an equality.
$endgroup$
– Did
Jan 13 at 20:46
add a comment |
$begingroup$
X ~ Bernoulli($p$) and Y ~ Poisson($p$) where $0<p<1$ and $p$ is the fixed number.
What is the maximum value of $P(X=Y)$ where the maximum value is taken over all possible joint distributions with specified marginal distributions?
That is... what is the joint distribution that maximizes the probability of X being equal to Y?
If I assume that $X$ and $Y$ are independent, finding the probability is easy since the only possible value $X$ can be is either $0$ or $1$. However, if I don't assume their independence, things get complicated. Any hints or help please?
probability probability-theory probability-distributions
edited Jan 13 at 20:43
Bernard
119k740113
asked Jan 13 at 20:24
JHKJHK
12
$endgroup$
X ~ Bernoulli($p$) and Y ~ Poisson($p$) where $0<p<1$ and $p$ is the fixed number.
What is the maximum value of $P(X=Y)$ where the maximum value is taken over all possible joint distributions with specified marginal distributions?
That is... what is the joint distribution that maximizes the probability of X being equal to Y?
If I assume that $X$ and $Y$ are independent, finding the probability is easy since the only possible value $X$ can be is either $0$ or $1$. However, if I don't assume their independence, things get complicated. Any hints or help please?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
edited Jan 13 at 20:43
Bernard
119k740113
asked Jan 13 at 20:24
JHKJHK
12
edited Jan 13 at 20:43
Bernard
119k740113
asked Jan 13 at 20:24
JHKJHK
12
edited Jan 13 at 20:43
Bernard
119k740113
edited Jan 13 at 20:43
Bernard
119k740113
edited Jan 13 at 20:43
Bernard
119k740113
119k740113
asked Jan 13 at 20:24
JHKJHK
12
asked Jan 13 at 20:24
JHKJHK
12
asked Jan 13 at 20:24
JHKJHK
12
12
$begingroup$
Since $P(X=0)=1-p<e^{-p}=P(Y=0)$, $P(X=1)=p>pe^{-p}=P(Y=1)$, and $P(Xgeqslant2)=0$, the maximum is $$1-p+pe^{-p}$$
$endgroup$
– Did
Jan 13 at 20:30
$begingroup$
@Did That comment should be an answer, I think
$endgroup$
– leonbloy
Jan 13 at 20:31
$begingroup$
@Did Are we adding the lower probability for each case to ensure the equality? like.. adding the upper bound for each case?
$endgroup$
– JHK
Jan 13 at 20:45
$begingroup$
Use the remark that $$P(X=Y)=sum_nP(X=n,Y=n)leqslantsum_nmin{P(X=n),P(Y=n)}$$ hence in the present case, $$P(X=Y)leqslant P(X=0)+P(Y=1)$$ and to complete the proof, it remains to build $(X,Y)$ such that the inequality is an equality.
$endgroup$
– Did
Jan 13 at 20:46
add a comment |
$begingroup$
Since $P(X=0)=1-p<e^{-p}=P(Y=0)$, $P(X=1)=p>pe^{-p}=P(Y=1)$, and $P(Xgeqslant2)=0$, the maximum is $$1-p+pe^{-p}$$
$endgroup$
– Did
Jan 13 at 20:30
$begingroup$
@Did That comment should be an answer, I think
$endgroup$
– leonbloy
Jan 13 at 20:31
$begingroup$
@Did Are we adding the lower probability for each case to ensure the equality? like.. adding the upper bound for each case?
$endgroup$
– JHK
Jan 13 at 20:45
$begingroup$
Use the remark that $$P(X=Y)=sum_nP(X=n,Y=n)leqslantsum_nmin{P(X=n),P(Y=n)}$$ hence in the present case, $$P(X=Y)leqslant P(X=0)+P(Y=1)$$ and to complete the proof, it remains to build $(X,Y)$ such that the inequality is an equality.
$endgroup$
– Did
Jan 13 at 20:46
$begingroup$
Since $P(X=0)=1-p<e^{-p}=P(Y=0)$, $P(X=1)=p>pe^{-p}=P(Y=1)$, and $P(Xgeqslant2)=0$, the maximum is $$1-p+pe^{-p}$$
$endgroup$
– Did
Jan 13 at 20:30
$begingroup$
Since $P(X=0)=1-p<e^{-p}=P(Y=0)$, $P(X=1)=p>pe^{-p}=P(Y=1)$, and $P(Xgeqslant2)=0$, the maximum is $$1-p+pe^{-p}$$
$endgroup$
– Did
Jan 13 at 20:30
$begingroup$
@Did That comment should be an answer, I think
$endgroup$
– leonbloy
Jan 13 at 20:31
$begingroup$
@Did That comment should be an answer, I think
$endgroup$
– leonbloy
Jan 13 at 20:31
$begingroup$
@Did Are we adding the lower probability for each case to ensure the equality? like.. adding the upper bound for each case?
$endgroup$
– JHK
Jan 13 at 20:45
$begingroup$
@Did Are we adding the lower probability for each case to ensure the equality? like.. adding the upper bound for each case?
$endgroup$
– JHK
Jan 13 at 20:45
$begingroup$
Use the remark that $$P(X=Y)=sum_nP(X=n,Y=n)leqslantsum_nmin{P(X=n),P(Y=n)}$$ hence in the present case, $$P(X=Y)leqslant P(X=0)+P(Y=1)$$ and to complete the proof, it remains to build $(X,Y)$ such that the inequality is an equality.
$endgroup$
– Did
Jan 13 at 20:46
$begingroup$
Use the remark that $$P(X=Y)=sum_nP(X=n,Y=n)leqslantsum_nmin{P(X=n),P(Y=n)}$$ hence in the present case, $$P(X=Y)leqslant P(X=0)+P(Y=1)$$ and to complete the proof, it remains to build $(X,Y)$ such that the inequality is an equality.
$endgroup$
– Did
Jan 13 at 20:46
add a comment |
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$begingroup$
Since $P(X=0)=1-p<e^{-p}=P(Y=0)$, $P(X=1)=p>pe^{-p}=P(Y=1)$, and $P(Xgeqslant2)=0$, the maximum is $$1-p+pe^{-p}$$
$endgroup$
– Did
Jan 13 at 20:30
$begingroup$
@Did That comment should be an answer, I think
$endgroup$
– leonbloy
Jan 13 at 20:31
$begingroup$
@Did Are we adding the lower probability for each case to ensure the equality? like.. adding the upper bound for each case?
$endgroup$
– JHK
Jan 13 at 20:45
$begingroup$
Use the remark that $$P(X=Y)=sum_nP(X=n,Y=n)leqslantsum_nmin{P(X=n),P(Y=n)}$$ hence in the present case, $$P(X=Y)leqslant P(X=0)+P(Y=1)$$ and to complete the proof, it remains to build $(X,Y)$ such that the inequality is an equality.
$endgroup$
– Did
Jan 13 at 20:46