How do I compute P(X=0)? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Help with Probability distributionsA Question with both poisson and gamma distributionpoisson distribution problem, expected value and variance calculationApplication Problem: Random Sums of Random Variables and CorrelationFind the variance of Y .Statistics: Deriving a Joint Probability Function From a Definition of Other PDF'sProbability: Finding the Expected Value of a Random Variable Derived From a Definition.Expected value of Poisson DistributionPoisson distribution cost functionPoisson parameter 𝜇 and expected occurrence of an event
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How do I compute P(X=0)?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Help with Probability distributionsA Question with both poisson and gamma distributionpoisson distribution problem, expected value and variance calculationApplication Problem: Random Sums of Random Variables and CorrelationFind the variance of Y .Statistics: Deriving a Joint Probability Function From a Definition of Other PDF'sProbability: Finding the Expected Value of a Random Variable Derived From a Definition.Expected value of Poisson DistributionPoisson distribution cost functionPoisson parameter 𝜇 and expected occurrence of an event
$begingroup$
A marine biology research group is looking for a kind of precious aquatic plant in a particular sea area. Let X be the number of the plants per cubic kilometer in the sea area and assume X has the Poisson distribution Poisson(λ). Also, the parameter λ varies with the location and has a Gamma distribution with parameters α and β.
(a) What is the expected number of plants per cubic kilometer?
(b) What is the variance of the number of plants per cubic kilometer?
(c) What is the probability they can find at least one plant per cubic kilometer?
For 1 c)
I got up to P(X>= 1) = 1- P(X=0) but am unsure of how to compute P(X=0)
probability statistics
asked Mar 27 at 17:24
RitaRita
64
$endgroup$
add a comment |
$begingroup$
A marine biology research group is looking for a kind of precious aquatic plant in a particular sea area. Let X be the number of the plants per cubic kilometer in the sea area and assume X has the Poisson distribution Poisson(λ). Also, the parameter λ varies with the location and has a Gamma distribution with parameters α and β.
(a) What is the expected number of plants per cubic kilometer?
(b) What is the variance of the number of plants per cubic kilometer?
(c) What is the probability they can find at least one plant per cubic kilometer?
For 1 c)
I got up to P(X>= 1) = 1- P(X=0) but am unsure of how to compute P(X=0)
probability statistics
asked Mar 27 at 17:24
RitaRita
64
$endgroup$
$begingroup$
Not sure about the scope of the question but when $lambda$ is constant, since $X sim mathcalP(lambda)$, you have $$ mathbbP[X=0] = e^-lambda fraclambda^00!=e^-lambda. $$
$endgroup$
– gt6989b
Mar 27 at 18:09
add a comment |
$begingroup$
A marine biology research group is looking for a kind of precious aquatic plant in a particular sea area. Let X be the number of the plants per cubic kilometer in the sea area and assume X has the Poisson distribution Poisson(λ). Also, the parameter λ varies with the location and has a Gamma distribution with parameters α and β.
(a) What is the expected number of plants per cubic kilometer?
(b) What is the variance of the number of plants per cubic kilometer?
(c) What is the probability they can find at least one plant per cubic kilometer?
For 1 c)
I got up to P(X>= 1) = 1- P(X=0) but am unsure of how to compute P(X=0)
probability statistics
asked Mar 27 at 17:24
RitaRita
64
$endgroup$
A marine biology research group is looking for a kind of precious aquatic plant in a particular sea area. Let X be the number of the plants per cubic kilometer in the sea area and assume X has the Poisson distribution Poisson(λ). Also, the parameter λ varies with the location and has a Gamma distribution with parameters α and β.
(a) What is the expected number of plants per cubic kilometer?
(b) What is the variance of the number of plants per cubic kilometer?
(c) What is the probability they can find at least one plant per cubic kilometer?
For 1 c)
I got up to P(X>= 1) = 1- P(X=0) but am unsure of how to compute P(X=0)
probability statistics
probability statistics
asked Mar 27 at 17:24
RitaRita
64
asked Mar 27 at 17:24
RitaRita
64
asked Mar 27 at 17:24
RitaRita
64
asked Mar 27 at 17:24
RitaRita
64
asked Mar 27 at 17:24
RitaRita
64
64
$begingroup$
Not sure about the scope of the question but when $lambda$ is constant, since $X sim mathcalP(lambda)$, you have $$ mathbbP[X=0] = e^-lambda fraclambda^00!=e^-lambda. $$
$endgroup$
– gt6989b
Mar 27 at 18:09
add a comment |
$begingroup$
Not sure about the scope of the question but when $lambda$ is constant, since $X sim mathcalP(lambda)$, you have $$ mathbbP[X=0] = e^-lambda fraclambda^00!=e^-lambda. $$
$endgroup$
– gt6989b
Mar 27 at 18:09
$begingroup$
Not sure about the scope of the question but when $lambda$ is constant, since $X sim mathcalP(lambda)$, you have $$ mathbbP[X=0] = e^-lambda fraclambda^00!=e^-lambda. $$
$endgroup$
– gt6989b
Mar 27 at 18:09
$begingroup$
Not sure about the scope of the question but when $lambda$ is constant, since $X sim mathcalP(lambda)$, you have $$ mathbbP[X=0] = e^-lambda fraclambda^00!=e^-lambda. $$
$endgroup$
– gt6989b
Mar 27 at 18:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'll try to answer you question partially.
The idea behind the problem is the following.
If you exactly know $lambda$, you can say for sure that expected number of plants per cubic kilometer is $lambda$.
However you don't exactly know $lambda$, so the expected number of plants can be computed using the conditional probabilities. $E(X|lambda = hatlambda)= hatlambda$.
$$E(X) = E(E(X|lambda = hatlambda))$$
So you should take average over all possible averages (i.e. for all possible values of $lambda$ counting(or averaging) of course with the pdf of $lambda$).
So you will have
$$E(X) = E(E(X|lambda = hatlambda))=int_0^inftyhatlambdaf_lambda(hatlambda)dhatlambda=int_0^inftyhatlambdafracbeta^alphaGamma(alpha)hatlambda^alpha -1e^-beta hatlambdadhatlambda$$.
To our luck this is exactly the expectation of any gamma distribution. Which is $fracalphabeta$.
For the variance and probability the idea in identical. (Try yourself using the comment above.).
However for more rigorous treatment of the problem look Poisson point processes, (This is another huge universe.) as your case is spatial.
Sorry if i didn't help.
answered Mar 27 at 18:31
kolobokishkolobokish
41139
$endgroup$
add a comment |
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$begingroup$
I'll try to answer you question partially.
The idea behind the problem is the following.
If you exactly know $lambda$, you can say for sure that expected number of plants per cubic kilometer is $lambda$.
However you don't exactly know $lambda$, so the expected number of plants can be computed using the conditional probabilities. $E(X|lambda = hatlambda)= hatlambda$.
$$E(X) = E(E(X|lambda = hatlambda))$$
So you should take average over all possible averages (i.e. for all possible values of $lambda$ counting(or averaging) of course with the pdf of $lambda$).
So you will have
$$E(X) = E(E(X|lambda = hatlambda))=int_0^inftyhatlambdaf_lambda(hatlambda)dhatlambda=int_0^inftyhatlambdafracbeta^alphaGamma(alpha)hatlambda^alpha -1e^-beta hatlambdadhatlambda$$.
To our luck this is exactly the expectation of any gamma distribution. Which is $fracalphabeta$.
For the variance and probability the idea in identical. (Try yourself using the comment above.).
However for more rigorous treatment of the problem look Poisson point processes, (This is another huge universe.) as your case is spatial.
Sorry if i didn't help.
answered Mar 27 at 18:31
kolobokishkolobokish
41139
$endgroup$
add a comment |
$begingroup$
I'll try to answer you question partially.
The idea behind the problem is the following.
If you exactly know $lambda$, you can say for sure that expected number of plants per cubic kilometer is $lambda$.
However you don't exactly know $lambda$, so the expected number of plants can be computed using the conditional probabilities. $E(X|lambda = hatlambda)= hatlambda$.
$$E(X) = E(E(X|lambda = hatlambda))$$
So you should take average over all possible averages (i.e. for all possible values of $lambda$ counting(or averaging) of course with the pdf of $lambda$).
So you will have
$$E(X) = E(E(X|lambda = hatlambda))=int_0^inftyhatlambdaf_lambda(hatlambda)dhatlambda=int_0^inftyhatlambdafracbeta^alphaGamma(alpha)hatlambda^alpha -1e^-beta hatlambdadhatlambda$$.
To our luck this is exactly the expectation of any gamma distribution. Which is $fracalphabeta$.
For the variance and probability the idea in identical. (Try yourself using the comment above.).
However for more rigorous treatment of the problem look Poisson point processes, (This is another huge universe.) as your case is spatial.
Sorry if i didn't help.
answered Mar 27 at 18:31
kolobokishkolobokish
41139
$endgroup$
add a comment |
$begingroup$
I'll try to answer you question partially.
The idea behind the problem is the following.
If you exactly know $lambda$, you can say for sure that expected number of plants per cubic kilometer is $lambda$.
However you don't exactly know $lambda$, so the expected number of plants can be computed using the conditional probabilities. $E(X|lambda = hatlambda)= hatlambda$.
$$E(X) = E(E(X|lambda = hatlambda))$$
So you should take average over all possible averages (i.e. for all possible values of $lambda$ counting(or averaging) of course with the pdf of $lambda$).
So you will have
$$E(X) = E(E(X|lambda = hatlambda))=int_0^inftyhatlambdaf_lambda(hatlambda)dhatlambda=int_0^inftyhatlambdafracbeta^alphaGamma(alpha)hatlambda^alpha -1e^-beta hatlambdadhatlambda$$.
To our luck this is exactly the expectation of any gamma distribution. Which is $fracalphabeta$.
For the variance and probability the idea in identical. (Try yourself using the comment above.).
However for more rigorous treatment of the problem look Poisson point processes, (This is another huge universe.) as your case is spatial.
Sorry if i didn't help.
answered Mar 27 at 18:31
kolobokishkolobokish
41139
$endgroup$
I'll try to answer you question partially.
The idea behind the problem is the following.
If you exactly know $lambda$, you can say for sure that expected number of plants per cubic kilometer is $lambda$.
However you don't exactly know $lambda$, so the expected number of plants can be computed using the conditional probabilities. $E(X|lambda = hatlambda)= hatlambda$.
$$E(X) = E(E(X|lambda = hatlambda))$$
So you should take average over all possible averages (i.e. for all possible values of $lambda$ counting(or averaging) of course with the pdf of $lambda$).
So you will have
$$E(X) = E(E(X|lambda = hatlambda))=int_0^inftyhatlambdaf_lambda(hatlambda)dhatlambda=int_0^inftyhatlambdafracbeta^alphaGamma(alpha)hatlambda^alpha -1e^-beta hatlambdadhatlambda$$.
To our luck this is exactly the expectation of any gamma distribution. Which is $fracalphabeta$.
For the variance and probability the idea in identical. (Try yourself using the comment above.).
However for more rigorous treatment of the problem look Poisson point processes, (This is another huge universe.) as your case is spatial.
Sorry if i didn't help.
answered Mar 27 at 18:31
kolobokishkolobokish
41139
answered Mar 27 at 18:31
kolobokishkolobokish
41139
answered Mar 27 at 18:31
kolobokishkolobokish
41139
answered Mar 27 at 18:31
kolobokishkolobokish
41139
41139
add a comment |
add a comment |
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$begingroup$
Not sure about the scope of the question but when $lambda$ is constant, since $X sim mathcalP(lambda)$, you have $$ mathbbP[X=0] = e^-lambda fraclambda^00!=e^-lambda. $$
$endgroup$
– gt6989b
Mar 27 at 18:09