Probability Density function of Poisson distribution












1












$begingroup$


This is an assignment I got for my course on Stochastic Processes:




Let us consider a random variable X distributed as a Poisson P (λ)
where λ ∼ [0.5, 1].



(a) Which are the unconditional mean and variance for variable X? (DONE)



(b) Which is the probability density function of X? (Not need to solve
the integral)




I managed to do the first part (a) but the second part (b) doesn't make sense to me.



How can X have a probability density function if X is a Poisson and the poisson is discrete?
Am I missing something?
Also, the professor says that there is no need to solve the integral, but how can there be an integral if the Poisson is discrete?



I guess the answer lies in this part:




λ ∼ [0.5, 1]




But I can't find it.



This is what I did to solve (a)






self-study poisson-distribution stochastic-processes poisson-process







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Xi'an I don't know, she didn't add anything to that.
    $endgroup$
    – Edoardo Busetti
    20 hours ago










  • $begingroup$
    @Xi'an Could you please explain me how to do that? For question (a) I simply demonstrated that the mean and variance of a poisson is given by λ.
    $endgroup$
    – Edoardo Busetti
    20 hours ago






  • 1




    $begingroup$
    I suspect the vague notation "$lambdasim[0.5,1]$" might be intended to stipulate that $lambda$ is a random variable with a uniform distribution on the interval $[0.5,1].$
    $endgroup$
    – whuber♦
    17 hours ago
















1












$begingroup$


This is an assignment I got for my course on Stochastic Processes:




Let us consider a random variable X distributed as a Poisson P (λ)
where λ ∼ [0.5, 1].



(a) Which are the unconditional mean and variance for variable X? (DONE)



(b) Which is the probability density function of X? (Not need to solve
the integral)




I managed to do the first part (a) but the second part (b) doesn't make sense to me.



How can X have a probability density function if X is a Poisson and the poisson is discrete?
Am I missing something?
Also, the professor says that there is no need to solve the integral, but how can there be an integral if the Poisson is discrete?



I guess the answer lies in this part:




λ ∼ [0.5, 1]




But I can't find it.



This is what I did to solve (a)






self-study poisson-distribution stochastic-processes poisson-process







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Xi'an I don't know, she didn't add anything to that.
    $endgroup$
    – Edoardo Busetti
    20 hours ago










  • $begingroup$
    @Xi'an Could you please explain me how to do that? For question (a) I simply demonstrated that the mean and variance of a poisson is given by λ.
    $endgroup$
    – Edoardo Busetti
    20 hours ago






  • 1




    $begingroup$
    I suspect the vague notation "$lambdasim[0.5,1]$" might be intended to stipulate that $lambda$ is a random variable with a uniform distribution on the interval $[0.5,1].$
    $endgroup$
    – whuber♦
    17 hours ago














1












1








1





$begingroup$


This is an assignment I got for my course on Stochastic Processes:




Let us consider a random variable X distributed as a Poisson P (λ)
where λ ∼ [0.5, 1].



(a) Which are the unconditional mean and variance for variable X? (DONE)



(b) Which is the probability density function of X? (Not need to solve
the integral)




I managed to do the first part (a) but the second part (b) doesn't make sense to me.



How can X have a probability density function if X is a Poisson and the poisson is discrete?
Am I missing something?
Also, the professor says that there is no need to solve the integral, but how can there be an integral if the Poisson is discrete?



I guess the answer lies in this part:




λ ∼ [0.5, 1]




But I can't find it.



This is what I did to solve (a)






self-study poisson-distribution stochastic-processes poisson-process







share|cite|improve this question











$endgroup$




This is an assignment I got for my course on Stochastic Processes:




Let us consider a random variable X distributed as a Poisson P (λ)
where λ ∼ [0.5, 1].



(a) Which are the unconditional mean and variance for variable X? (DONE)



(b) Which is the probability density function of X? (Not need to solve
the integral)




I managed to do the first part (a) but the second part (b) doesn't make sense to me.



How can X have a probability density function if X is a Poisson and the poisson is discrete?
Am I missing something?
Also, the professor says that there is no need to solve the integral, but how can there be an integral if the Poisson is discrete?



I guess the answer lies in this part:




λ ∼ [0.5, 1]




But I can't find it.



This is what I did to solve (a)






self-study poisson-distribution stochastic-processes poisson-process




self-study poisson-distribution stochastic-processes poisson-process






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 20 hours ago







Edoardo Busetti

















asked 20 hours ago









Edoardo BusettiEdoardo Busetti

475




475












  • $begingroup$
    @Xi'an I don't know, she didn't add anything to that.
    $endgroup$
    – Edoardo Busetti
    20 hours ago










  • $begingroup$
    @Xi'an Could you please explain me how to do that? For question (a) I simply demonstrated that the mean and variance of a poisson is given by λ.
    $endgroup$
    – Edoardo Busetti
    20 hours ago






  • 1




    $begingroup$
    I suspect the vague notation "$lambdasim[0.5,1]$" might be intended to stipulate that $lambda$ is a random variable with a uniform distribution on the interval $[0.5,1].$
    $endgroup$
    – whuber♦
    17 hours ago


















  • $begingroup$
    @Xi'an I don't know, she didn't add anything to that.
    $endgroup$
    – Edoardo Busetti
    20 hours ago










  • $begingroup$
    @Xi'an Could you please explain me how to do that? For question (a) I simply demonstrated that the mean and variance of a poisson is given by λ.
    $endgroup$
    – Edoardo Busetti
    20 hours ago






  • 1




    $begingroup$
    I suspect the vague notation "$lambdasim[0.5,1]$" might be intended to stipulate that $lambda$ is a random variable with a uniform distribution on the interval $[0.5,1].$
    $endgroup$
    – whuber♦
    17 hours ago
















$begingroup$
@Xi'an I don't know, she didn't add anything to that.
$endgroup$
– Edoardo Busetti
20 hours ago




$begingroup$
@Xi'an I don't know, she didn't add anything to that.
$endgroup$
– Edoardo Busetti
20 hours ago












$begingroup$
@Xi'an Could you please explain me how to do that? For question (a) I simply demonstrated that the mean and variance of a poisson is given by λ.
$endgroup$
– Edoardo Busetti
20 hours ago




$begingroup$
@Xi'an Could you please explain me how to do that? For question (a) I simply demonstrated that the mean and variance of a poisson is given by λ.
$endgroup$
– Edoardo Busetti
20 hours ago




1




1




$begingroup$
I suspect the vague notation "$lambdasim[0.5,1]$" might be intended to stipulate that $lambda$ is a random variable with a uniform distribution on the interval $[0.5,1].$
$endgroup$
– whuber♦
17 hours ago




$begingroup$
I suspect the vague notation "$lambdasim[0.5,1]$" might be intended to stipulate that $lambda$ is a random variable with a uniform distribution on the interval $[0.5,1].$
$endgroup$
– whuber♦
17 hours ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

Question (a): One need call the Law of Total Expectation and Law of Total Variance.




$$mathbb{E}(X)=mathbb{E}[mathbb{E}(Xmid lambda)])$$
where $mathbb{E}(Xmid lambda)=lambda$ and
$$operatorname{Var}(X)=mathbb{E}[operatorname{Var}(Xmid lambda)] + operatorname{Var}(mathbb{E}[Xmid lambda])$$
where $operatorname{Var}(Xmid lambda)=lambda$




Question (b): If the integral need not be solved, the marginal density of $X$ can be written as the integral (when $kinmathbb{N}$)




$$p(k)=2int_{0.5}^1 frac{lambda^k}{k!}e^{-lambda},text{d}lambda$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm having trouble understanding how to calculate Var(X)=𝔼[Var(X∣λ)]+Var(𝔼[X∣λ]). I think it is: Var(X)=𝔼[Var(X∣λ)] + Var(λ). Where the Var(λ) is the variance of an uniform distribution, but I don't understand how to calculate 𝔼[Var(X∣λ)].
    $endgroup$
    – Edoardo Busetti
    19 hours ago










  • $begingroup$
    $text{var}(X|lambda)$ is the variance of the Poisson $P(lambda)$ distribution.
    $endgroup$
    – Xi'an
    17 hours ago











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1 Answer
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1 Answer
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active

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3












$begingroup$

Question (a): One need call the Law of Total Expectation and Law of Total Variance.




$$mathbb{E}(X)=mathbb{E}[mathbb{E}(Xmid lambda)])$$
where $mathbb{E}(Xmid lambda)=lambda$ and
$$operatorname{Var}(X)=mathbb{E}[operatorname{Var}(Xmid lambda)] + operatorname{Var}(mathbb{E}[Xmid lambda])$$
where $operatorname{Var}(Xmid lambda)=lambda$




Question (b): If the integral need not be solved, the marginal density of $X$ can be written as the integral (when $kinmathbb{N}$)




$$p(k)=2int_{0.5}^1 frac{lambda^k}{k!}e^{-lambda},text{d}lambda$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm having trouble understanding how to calculate Var(X)=𝔼[Var(X∣λ)]+Var(𝔼[X∣λ]). I think it is: Var(X)=𝔼[Var(X∣λ)] + Var(λ). Where the Var(λ) is the variance of an uniform distribution, but I don't understand how to calculate 𝔼[Var(X∣λ)].
    $endgroup$
    – Edoardo Busetti
    19 hours ago










  • $begingroup$
    $text{var}(X|lambda)$ is the variance of the Poisson $P(lambda)$ distribution.
    $endgroup$
    – Xi'an
    17 hours ago
















3












$begingroup$

Question (a): One need call the Law of Total Expectation and Law of Total Variance.




$$mathbb{E}(X)=mathbb{E}[mathbb{E}(Xmid lambda)])$$
where $mathbb{E}(Xmid lambda)=lambda$ and
$$operatorname{Var}(X)=mathbb{E}[operatorname{Var}(Xmid lambda)] + operatorname{Var}(mathbb{E}[Xmid lambda])$$
where $operatorname{Var}(Xmid lambda)=lambda$




Question (b): If the integral need not be solved, the marginal density of $X$ can be written as the integral (when $kinmathbb{N}$)




$$p(k)=2int_{0.5}^1 frac{lambda^k}{k!}e^{-lambda},text{d}lambda$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm having trouble understanding how to calculate Var(X)=𝔼[Var(X∣λ)]+Var(𝔼[X∣λ]). I think it is: Var(X)=𝔼[Var(X∣λ)] + Var(λ). Where the Var(λ) is the variance of an uniform distribution, but I don't understand how to calculate 𝔼[Var(X∣λ)].
    $endgroup$
    – Edoardo Busetti
    19 hours ago










  • $begingroup$
    $text{var}(X|lambda)$ is the variance of the Poisson $P(lambda)$ distribution.
    $endgroup$
    – Xi'an
    17 hours ago














3












3








3





$begingroup$

Question (a): One need call the Law of Total Expectation and Law of Total Variance.




$$mathbb{E}(X)=mathbb{E}[mathbb{E}(Xmid lambda)])$$
where $mathbb{E}(Xmid lambda)=lambda$ and
$$operatorname{Var}(X)=mathbb{E}[operatorname{Var}(Xmid lambda)] + operatorname{Var}(mathbb{E}[Xmid lambda])$$
where $operatorname{Var}(Xmid lambda)=lambda$




Question (b): If the integral need not be solved, the marginal density of $X$ can be written as the integral (when $kinmathbb{N}$)




$$p(k)=2int_{0.5}^1 frac{lambda^k}{k!}e^{-lambda},text{d}lambda$$







share|cite|improve this answer











$endgroup$



Question (a): One need call the Law of Total Expectation and Law of Total Variance.




$$mathbb{E}(X)=mathbb{E}[mathbb{E}(Xmid lambda)])$$
where $mathbb{E}(Xmid lambda)=lambda$ and
$$operatorname{Var}(X)=mathbb{E}[operatorname{Var}(Xmid lambda)] + operatorname{Var}(mathbb{E}[Xmid lambda])$$
where $operatorname{Var}(Xmid lambda)=lambda$




Question (b): If the integral need not be solved, the marginal density of $X$ can be written as the integral (when $kinmathbb{N}$)




$$p(k)=2int_{0.5}^1 frac{lambda^k}{k!}e^{-lambda},text{d}lambda$$








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 17 hours ago

























answered 20 hours ago









Xi'anXi'an

54.6k692351




54.6k692351












  • $begingroup$
    I'm having trouble understanding how to calculate Var(X)=𝔼[Var(X∣λ)]+Var(𝔼[X∣λ]). I think it is: Var(X)=𝔼[Var(X∣λ)] + Var(λ). Where the Var(λ) is the variance of an uniform distribution, but I don't understand how to calculate 𝔼[Var(X∣λ)].
    $endgroup$
    – Edoardo Busetti
    19 hours ago










  • $begingroup$
    $text{var}(X|lambda)$ is the variance of the Poisson $P(lambda)$ distribution.
    $endgroup$
    – Xi'an
    17 hours ago


















  • $begingroup$
    I'm having trouble understanding how to calculate Var(X)=𝔼[Var(X∣λ)]+Var(𝔼[X∣λ]). I think it is: Var(X)=𝔼[Var(X∣λ)] + Var(λ). Where the Var(λ) is the variance of an uniform distribution, but I don't understand how to calculate 𝔼[Var(X∣λ)].
    $endgroup$
    – Edoardo Busetti
    19 hours ago










  • $begingroup$
    $text{var}(X|lambda)$ is the variance of the Poisson $P(lambda)$ distribution.
    $endgroup$
    – Xi'an
    17 hours ago
















$begingroup$
I'm having trouble understanding how to calculate Var(X)=𝔼[Var(X∣λ)]+Var(𝔼[X∣λ]). I think it is: Var(X)=𝔼[Var(X∣λ)] + Var(λ). Where the Var(λ) is the variance of an uniform distribution, but I don't understand how to calculate 𝔼[Var(X∣λ)].
$endgroup$
– Edoardo Busetti
19 hours ago




$begingroup$
I'm having trouble understanding how to calculate Var(X)=𝔼[Var(X∣λ)]+Var(𝔼[X∣λ]). I think it is: Var(X)=𝔼[Var(X∣λ)] + Var(λ). Where the Var(λ) is the variance of an uniform distribution, but I don't understand how to calculate 𝔼[Var(X∣λ)].
$endgroup$
– Edoardo Busetti
19 hours ago












$begingroup$
$text{var}(X|lambda)$ is the variance of the Poisson $P(lambda)$ distribution.
$endgroup$
– Xi'an
17 hours ago




$begingroup$
$text{var}(X|lambda)$ is the variance of the Poisson $P(lambda)$ distribution.
$endgroup$
– Xi'an
17 hours ago


















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