Problem on Poisson distribution
$begingroup$
Let $X$ be a discrete random variable with Poisson distribution
$$ p_X(k) = e^{-lambda} . lambda^k / k!$$ for $k=0,1,2,... $
If it's given that
$$p_X(i) = p_X(j) tag {*} $$
for some different integers $i < j$,
can we prove that
(1) $lambda$ is also an integer
(2) $i=lambda-1$ and $j=lambda$ are the only two indexes with the (*) property
I am reading some proof but it's very short and not very convincing at 1-2 points.
So I wanted to check with the community here.
UPDATE:
The proof which I have first obtains that:
$lambda^k = (i+1)(i+2)...(i+k)$
where $k=j-i$. This is easy to obtain from (*).
So far so good.
Then from there on it says:
"But this is only possible if k=1 (can you guess why?)"
Btw, I cannot guess why, not right now.
And from $k=1$ of course it follows trivially that
$i+1 = lambda$
which means that
$j= lambda$ and $i = lambda-1$
So that's all I have in my hands. This sketch of a proof.
probability probability-theory statistics
asked Dec 23 '18 at 22:09
peter.petrovpeter.petrov
5,439821
$endgroup$
|
show 5 more comments
$begingroup$
Let $X$ be a discrete random variable with Poisson distribution
$$ p_X(k) = e^{-lambda} . lambda^k / k!$$ for $k=0,1,2,... $
If it's given that
$$p_X(i) = p_X(j) tag {*} $$
for some different integers $i < j$,
can we prove that
(1) $lambda$ is also an integer
(2) $i=lambda-1$ and $j=lambda$ are the only two indexes with the (*) property
I am reading some proof but it's very short and not very convincing at 1-2 points.
So I wanted to check with the community here.
UPDATE:
The proof which I have first obtains that:
$lambda^k = (i+1)(i+2)...(i+k)$
where $k=j-i$. This is easy to obtain from (*).
So far so good.
Then from there on it says:
"But this is only possible if k=1 (can you guess why?)"
Btw, I cannot guess why, not right now.
And from $k=1$ of course it follows trivially that
$i+1 = lambda$
which means that
$j= lambda$ and $i = lambda-1$
So that's all I have in my hands. This sketch of a proof.
probability probability-theory statistics
asked Dec 23 '18 at 22:09
peter.petrovpeter.petrov
5,439821
$endgroup$
$begingroup$
"it's very short" A good oint, right? "and not very convincing at 1-2 points" Which ones? Unless you tell us them, how can we help?
$endgroup$
– Did
Dec 23 '18 at 22:22
$begingroup$
What is short? I don't understand... the proof? It is not long indeed but if I have to rewrite it with MathJax, it will take me ages, I am not very good with MathJax. I just don't understand if "$lambda$ is integer" can be inferred from what is given or if it has to be given too. This is my ask basically.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:51
$begingroup$
Btw, the problem is very interesting (at least for me). I am not sure why somebody downvoted.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:57
$begingroup$
@Did I added some info. That's really all I have.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:21
$begingroup$
OK. This sounds like a bad MSE answer... Is this it? Then you could as well add a link to it...
$endgroup$
– Did
Dec 24 '18 at 8:00
|
show 5 more comments
$begingroup$
Let $X$ be a discrete random variable with Poisson distribution
$$ p_X(k) = e^{-lambda} . lambda^k / k!$$ for $k=0,1,2,... $
If it's given that
$$p_X(i) = p_X(j) tag {*} $$
for some different integers $i < j$,
can we prove that
(1) $lambda$ is also an integer
(2) $i=lambda-1$ and $j=lambda$ are the only two indexes with the (*) property
I am reading some proof but it's very short and not very convincing at 1-2 points.
So I wanted to check with the community here.
UPDATE:
The proof which I have first obtains that:
$lambda^k = (i+1)(i+2)...(i+k)$
where $k=j-i$. This is easy to obtain from (*).
So far so good.
Then from there on it says:
"But this is only possible if k=1 (can you guess why?)"
Btw, I cannot guess why, not right now.
And from $k=1$ of course it follows trivially that
$i+1 = lambda$
which means that
$j= lambda$ and $i = lambda-1$
So that's all I have in my hands. This sketch of a proof.
probability probability-theory statistics
asked Dec 23 '18 at 22:09
peter.petrovpeter.petrov
5,439821
$endgroup$
Let $X$ be a discrete random variable with Poisson distribution
$$ p_X(k) = e^{-lambda} . lambda^k / k!$$ for $k=0,1,2,... $
If it's given that
$$p_X(i) = p_X(j) tag {*} $$
for some different integers $i < j$,
can we prove that
(1) $lambda$ is also an integer
(2) $i=lambda-1$ and $j=lambda$ are the only two indexes with the (*) property
I am reading some proof but it's very short and not very convincing at 1-2 points.
So I wanted to check with the community here.
UPDATE:
The proof which I have first obtains that:
$lambda^k = (i+1)(i+2)...(i+k)$
where $k=j-i$. This is easy to obtain from (*).
So far so good.
Then from there on it says:
"But this is only possible if k=1 (can you guess why?)"
Btw, I cannot guess why, not right now.
And from $k=1$ of course it follows trivially that
$i+1 = lambda$
which means that
$j= lambda$ and $i = lambda-1$
So that's all I have in my hands. This sketch of a proof.
probability probability-theory statistics
probability probability-theory statistics
asked Dec 23 '18 at 22:09
peter.petrovpeter.petrov
5,439821
asked Dec 23 '18 at 22:09
peter.petrovpeter.petrov
5,439821
edited Dec 23 '18 at 23:50
peter.petrov
asked Dec 23 '18 at 22:09
peter.petrovpeter.petrov
5,439821
asked Dec 23 '18 at 22:09
peter.petrovpeter.petrov
5,439821
asked Dec 23 '18 at 22:09
peter.petrovpeter.petrov
5,439821
5,439821
$begingroup$
"it's very short" A good oint, right? "and not very convincing at 1-2 points" Which ones? Unless you tell us them, how can we help?
$endgroup$
– Did
Dec 23 '18 at 22:22
$begingroup$
What is short? I don't understand... the proof? It is not long indeed but if I have to rewrite it with MathJax, it will take me ages, I am not very good with MathJax. I just don't understand if "$lambda$ is integer" can be inferred from what is given or if it has to be given too. This is my ask basically.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:51
$begingroup$
Btw, the problem is very interesting (at least for me). I am not sure why somebody downvoted.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:57
$begingroup$
@Did I added some info. That's really all I have.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:21
$begingroup$
OK. This sounds like a bad MSE answer... Is this it? Then you could as well add a link to it...
$endgroup$
– Did
Dec 24 '18 at 8:00
|
show 5 more comments
$begingroup$
"it's very short" A good oint, right? "and not very convincing at 1-2 points" Which ones? Unless you tell us them, how can we help?
$endgroup$
– Did
Dec 23 '18 at 22:22
$begingroup$
What is short? I don't understand... the proof? It is not long indeed but if I have to rewrite it with MathJax, it will take me ages, I am not very good with MathJax. I just don't understand if "$lambda$ is integer" can be inferred from what is given or if it has to be given too. This is my ask basically.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:51
$begingroup$
Btw, the problem is very interesting (at least for me). I am not sure why somebody downvoted.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:57
$begingroup$
@Did I added some info. That's really all I have.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:21
$begingroup$
OK. This sounds like a bad MSE answer... Is this it? Then you could as well add a link to it...
$endgroup$
– Did
Dec 24 '18 at 8:00
$begingroup$
"it's very short" A good oint, right? "and not very convincing at 1-2 points" Which ones? Unless you tell us them, how can we help?
$endgroup$
– Did
Dec 23 '18 at 22:22
$begingroup$
"it's very short" A good oint, right? "and not very convincing at 1-2 points" Which ones? Unless you tell us them, how can we help?
$endgroup$
– Did
Dec 23 '18 at 22:22
$begingroup$
What is short? I don't understand... the proof? It is not long indeed but if I have to rewrite it with MathJax, it will take me ages, I am not very good with MathJax. I just don't understand if "$lambda$ is integer" can be inferred from what is given or if it has to be given too. This is my ask basically.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:51
$begingroup$
What is short? I don't understand... the proof? It is not long indeed but if I have to rewrite it with MathJax, it will take me ages, I am not very good with MathJax. I just don't understand if "$lambda$ is integer" can be inferred from what is given or if it has to be given too. This is my ask basically.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:51
$begingroup$
Btw, the problem is very interesting (at least for me). I am not sure why somebody downvoted.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:57
$begingroup$
Btw, the problem is very interesting (at least for me). I am not sure why somebody downvoted.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:57
$begingroup$
@Did I added some info. That's really all I have.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:21
$begingroup$
@Did I added some info. That's really all I have.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:21
$begingroup$
OK. This sounds like a bad MSE answer... Is this it? Then you could as well add a link to it...
$endgroup$
– Did
Dec 24 '18 at 8:00
$begingroup$
OK. This sounds like a bad MSE answer... Is this it? Then you could as well add a link to it...
$endgroup$
– Did
Dec 24 '18 at 8:00
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
OK, I just found that we cannot infer (from what is given) that $lambda$ is an integer. So this should be given. But if it's given then... What is this all about? Isn't it all too easy?
I just constructed a counter example:
If $lambda = sqrt{6}$ then $P_X(1) = P_X(3)$, right?
I am really confused what this problem is about.
It is badly stated and badly proved. No idea.
That's why I asked my question, I was hoping someone would have an idea what this is all about.
answered Dec 23 '18 at 23:43
peter.petrovpeter.petrov
5,439821
$endgroup$
add a comment |
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$begingroup$
OK, I just found that we cannot infer (from what is given) that $lambda$ is an integer. So this should be given. But if it's given then... What is this all about? Isn't it all too easy?
I just constructed a counter example:
If $lambda = sqrt{6}$ then $P_X(1) = P_X(3)$, right?
I am really confused what this problem is about.
It is badly stated and badly proved. No idea.
That's why I asked my question, I was hoping someone would have an idea what this is all about.
answered Dec 23 '18 at 23:43
peter.petrovpeter.petrov
5,439821
$endgroup$
add a comment |
$begingroup$
OK, I just found that we cannot infer (from what is given) that $lambda$ is an integer. So this should be given. But if it's given then... What is this all about? Isn't it all too easy?
I just constructed a counter example:
If $lambda = sqrt{6}$ then $P_X(1) = P_X(3)$, right?
I am really confused what this problem is about.
It is badly stated and badly proved. No idea.
That's why I asked my question, I was hoping someone would have an idea what this is all about.
answered Dec 23 '18 at 23:43
peter.petrovpeter.petrov
5,439821
$endgroup$
add a comment |
$begingroup$
OK, I just found that we cannot infer (from what is given) that $lambda$ is an integer. So this should be given. But if it's given then... What is this all about? Isn't it all too easy?
I just constructed a counter example:
If $lambda = sqrt{6}$ then $P_X(1) = P_X(3)$, right?
I am really confused what this problem is about.
It is badly stated and badly proved. No idea.
That's why I asked my question, I was hoping someone would have an idea what this is all about.
answered Dec 23 '18 at 23:43
peter.petrovpeter.petrov
5,439821
$endgroup$
OK, I just found that we cannot infer (from what is given) that $lambda$ is an integer. So this should be given. But if it's given then... What is this all about? Isn't it all too easy?
I just constructed a counter example:
If $lambda = sqrt{6}$ then $P_X(1) = P_X(3)$, right?
I am really confused what this problem is about.
It is badly stated and badly proved. No idea.
That's why I asked my question, I was hoping someone would have an idea what this is all about.
answered Dec 23 '18 at 23:43
peter.petrovpeter.petrov
5,439821
answered Dec 23 '18 at 23:43
peter.petrovpeter.petrov
5,439821
answered Dec 23 '18 at 23:43
peter.petrovpeter.petrov
5,439821
answered Dec 23 '18 at 23:43
peter.petrovpeter.petrov
5,439821
5,439821
add a comment |
add a comment |
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$begingroup$
"it's very short" A good oint, right? "and not very convincing at 1-2 points" Which ones? Unless you tell us them, how can we help?
$endgroup$
– Did
Dec 23 '18 at 22:22
$begingroup$
What is short? I don't understand... the proof? It is not long indeed but if I have to rewrite it with MathJax, it will take me ages, I am not very good with MathJax. I just don't understand if "$lambda$ is integer" can be inferred from what is given or if it has to be given too. This is my ask basically.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:51
$begingroup$
Btw, the problem is very interesting (at least for me). I am not sure why somebody downvoted.
$endgroup$
– peter.petrov
Dec 23 '18 at 22:57
$begingroup$
@Did I added some info. That's really all I have.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:21
$begingroup$
OK. This sounds like a bad MSE answer... Is this it? Then you could as well add a link to it...
$endgroup$
– Did
Dec 24 '18 at 8:00