Poisson distribution - more than 2 points with same probability mass?!
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For a random variable X with Poisson distribution is it possible that:
$$p_X(i) = p_X(j) = p_X(s) tag {*} $$
for some different integers $i < j < s$?
I think it's not possible but... How can we (most simply) prove it?
probability probability-theory
asked Dec 23 '18 at 23:27
peter.petrovpeter.petrov
5,439821
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add a comment |
$begingroup$
For a random variable X with Poisson distribution is it possible that:
$$p_X(i) = p_X(j) = p_X(s) tag {*} $$
for some different integers $i < j < s$?
I think it's not possible but... How can we (most simply) prove it?
probability probability-theory
asked Dec 23 '18 at 23:27
peter.petrovpeter.petrov
5,439821
$endgroup$
$begingroup$
It is a consequence of the probability mass function being strictly increasing up to a particular value and then strictly decreasing from the next value
$endgroup$
– Henry
Dec 23 '18 at 23:35
add a comment |
$begingroup$
For a random variable X with Poisson distribution is it possible that:
$$p_X(i) = p_X(j) = p_X(s) tag {*} $$
for some different integers $i < j < s$?
I think it's not possible but... How can we (most simply) prove it?
probability probability-theory
asked Dec 23 '18 at 23:27
peter.petrovpeter.petrov
5,439821
$endgroup$
For a random variable X with Poisson distribution is it possible that:
$$p_X(i) = p_X(j) = p_X(s) tag {*} $$
for some different integers $i < j < s$?
I think it's not possible but... How can we (most simply) prove it?
probability probability-theory
probability probability-theory
asked Dec 23 '18 at 23:27
peter.petrovpeter.petrov
5,439821
asked Dec 23 '18 at 23:27
peter.petrovpeter.petrov
5,439821
asked Dec 23 '18 at 23:27
peter.petrovpeter.petrov
5,439821
asked Dec 23 '18 at 23:27
peter.petrovpeter.petrov
5,439821
asked Dec 23 '18 at 23:27
peter.petrovpeter.petrov
5,439821
5,439821
$begingroup$
It is a consequence of the probability mass function being strictly increasing up to a particular value and then strictly decreasing from the next value
$endgroup$
– Henry
Dec 23 '18 at 23:35
add a comment |
$begingroup$
It is a consequence of the probability mass function being strictly increasing up to a particular value and then strictly decreasing from the next value
$endgroup$
– Henry
Dec 23 '18 at 23:35
$begingroup$
It is a consequence of the probability mass function being strictly increasing up to a particular value and then strictly decreasing from the next value
$endgroup$
– Henry
Dec 23 '18 at 23:35
$begingroup$
It is a consequence of the probability mass function being strictly increasing up to a particular value and then strictly decreasing from the next value
$endgroup$
– Henry
Dec 23 '18 at 23:35
add a comment |
1 Answer
1
active
oldest
votes
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Hint
Let $p_{X}(k)=lambda^{k}e^{-lambda}/k!$. By considering $p_{X}(k+1)/p_{X}(k)$ show that $p_{X}$ is increasing for $kleq lambda-1$ and decreasing for $k>lambda-1$.
answered Dec 23 '18 at 23:35
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
$begingroup$
Thanks. You should say "strictly increasing/decreasing", I think.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:48
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Let $p_{X}(k)=lambda^{k}e^{-lambda}/k!$. By considering $p_{X}(k+1)/p_{X}(k)$ show that $p_{X}$ is increasing for $kleq lambda-1$ and decreasing for $k>lambda-1$.
answered Dec 23 '18 at 23:35
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
$begingroup$
Thanks. You should say "strictly increasing/decreasing", I think.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:48
add a comment |
$begingroup$
Hint
Let $p_{X}(k)=lambda^{k}e^{-lambda}/k!$. By considering $p_{X}(k+1)/p_{X}(k)$ show that $p_{X}$ is increasing for $kleq lambda-1$ and decreasing for $k>lambda-1$.
answered Dec 23 '18 at 23:35
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
$begingroup$
Thanks. You should say "strictly increasing/decreasing", I think.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:48
add a comment |
$begingroup$
Hint
Let $p_{X}(k)=lambda^{k}e^{-lambda}/k!$. By considering $p_{X}(k+1)/p_{X}(k)$ show that $p_{X}$ is increasing for $kleq lambda-1$ and decreasing for $k>lambda-1$.
answered Dec 23 '18 at 23:35
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
Hint
Let $p_{X}(k)=lambda^{k}e^{-lambda}/k!$. By considering $p_{X}(k+1)/p_{X}(k)$ show that $p_{X}$ is increasing for $kleq lambda-1$ and decreasing for $k>lambda-1$.
answered Dec 23 '18 at 23:35
Foobaz JohnFoobaz John
22.1k41352
answered Dec 23 '18 at 23:35
Foobaz JohnFoobaz John
22.1k41352
answered Dec 23 '18 at 23:35
Foobaz JohnFoobaz John
22.1k41352
answered Dec 23 '18 at 23:35
Foobaz JohnFoobaz John
22.1k41352
22.1k41352
$begingroup$
Thanks. You should say "strictly increasing/decreasing", I think.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:48
add a comment |
$begingroup$
Thanks. You should say "strictly increasing/decreasing", I think.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:48
$begingroup$
Thanks. You should say "strictly increasing/decreasing", I think.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:48
$begingroup$
Thanks. You should say "strictly increasing/decreasing", I think.
$endgroup$
– peter.petrov
Dec 23 '18 at 23:48
add a comment |
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It is a consequence of the probability mass function being strictly increasing up to a particular value and then strictly decreasing from the next value
$endgroup$
– Henry
Dec 23 '18 at 23:35