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Poisson distribution

suvadip

Member
Feb 21, 2013
69
If \(\displaystyle p(x=1)=p(x=2)\) where \(\displaystyle x\) follows a Poisson distribution, then find \(\displaystyle p(x=0 ~~or~~ 1) \). Also find \(\displaystyle F(x)\)


In connection with the above question, I have confusion about the last part i.e., about \(\displaystyle F(x)\). I can find \(\displaystyle E(x)\) here, but how to find \(\displaystyle F(x)\).
 

chisigma

Well-known member
Feb 13, 2012
1,704
If \(\displaystyle p(x=1)=p(x=2)\) where \(\displaystyle x\) follows a Poisson distribution, then find \(\displaystyle p(x=0 ~~or~~ 1) \). Also find \(\displaystyle F(x)\)


In connection with the above question, I have confusion about the last part i.e., about \(\displaystyle F(x)\). I can find \(\displaystyle E(x)\) here, but how to find \(\displaystyle F(x)\).
The Poisson distribution is associated with the probability function...

$\displaystyle P \{x=k\} = e^{- \lambda}\ \frac{\lambda^{k}}{k!}\ (1)$

... so that the condition $\displaystyle P \{x=1\} = P\{x=2\}$ is equivalent to say that $\displaystyle \lambda = \frac{\lambda^{2}}{2}$, and that happens for $\lambda = 0$ and $\lambda=2$...

Kind regards

$\chi$ $\sigma$
 

suvadip

Member
Feb 21, 2013
69
The Poisson distribution is associated with the probability function...

$\displaystyle P \{x=k\} = e^{- \lambda}\ \frac{\lambda^{k}}{k!}\ (1)$

... so that the condition $\displaystyle P \{x=1\} = P\{x=2\}$ is equivalent to say that $\displaystyle \lambda = \frac{\lambda^{2}}{2}$, and that happens for $\lambda = 0$ and $\lambda=2$...

Kind regards

$\chi$ $\sigma$

Upto that I have already done. I have confusion about how to find \(\displaystyle F(x)\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Upto that I have already done. I have confusion about how to find \(\displaystyle F(x)\)
The Poisson distribution is a discrete distribution, so that F(x) must be written in term of Heaviside Step Function as...

$\displaystyle F(x) = e^{- \lambda}\ \sum_{n = 0}^{\infty} \frac{\lambda^{n}}{n!}\ \mathcal {U} (x - n)\ (1)$

Kind regards

$\chi$ $\sigma$
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
The Poisson distribution is a discrete distribution, so that F(x) must be written in term of Heaviside Step Function as...

$\displaystyle F(x) = e^{- \lambda}\ \sum_{n = 0}^{\infty} \frac{\lambda^{n}}{n!}\ \mathcal {U} (x - n)\ (1)$

Kind regards

$\chi$ $\sigma$
I do not think he means that. His notation is confusing. He used $x$ for notation of a random variable instead of the more common $X$. In his question he says that he knows how to find $E(x)$. It seems that he is saying that he knows how to find the expectation of $x$, or in more standard notation $E[X]$. He is then asking how to find $F(x)$ which we can only assume he is saying $F(X)$ in more common notation, but I do not know what $F$ of a random variable is supposed to me, perhaps it is the variable of it.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Given that the problem involves some basic principles of the Poisson distribution and probability concepts, I think it's safe to assume that $F$ is the CDF for the random variable here. It makes sense that finding $E[X]$ was easy to do since for a Poisson rv there isn't any work to do but the CDF is slightly more complicated. Considering these things together, I fully agree with chisigma's interpretation.
 

suvadip

Member
Feb 21, 2013
69
Given that the problem involves some basic principles of the Poisson distribution and probability concepts, I think it's safe to assume that $F$ is the CDF for the random variable here. It makes sense that finding $E[X]$ was easy to do since for a Poisson rv there isn't any work to do but the CDF is slightly more complicated. Considering these things together, I fully agree with chisigma's interpretation.

\(\displaystyle F\) is the distribution function.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Do you mean this? If so, that's what I wrote. If it's not then maybe you are referring to the pmf. Which one is correct?

EDIT: Ok, let's try it this way so we aren't guessing what you know or don't know. What did you get when you tried to find $F(x)$? For continuous distributions we can simply integrate the pdf to find the cdf, but for discrete ones it can be trickier. In fact, the CDF for Poisson has no closed form as far as I know. Maybe this isn't what you are asked to find, but we need more info from you to proceed.
 

chisigma

Well-known member
Feb 13, 2012
1,704
I do not think he means that. His notation is confusing. He used $x$ for notation of a random variable instead of the more common $X$. In his question he says that he knows how to find $E(x)$. It seems that he is saying that he knows how to find the expectation of $x$, or in more standard notation $E[X]$. He is then asking how to find $F(x)$ which we can only assume he is saying $F(X)$ in more common notation, but I do not know what $F$ of a random variable is supposed to me, perhaps it is the variable of it.
In the language of probability F(x) usually indicates the Probability Distribution Function of a r.v. X, that by definition is...

$\displaystyle F(x) = P \{X \le x\}\ (1)$

Its derivative is the Probability Density Function and is indicated with f(x), so that is...

$\displaystyle F(x) = P \{X \le x\} = \int_{- \infty}^{x} f(\xi)\ d \xi\ (2)$

All that has no problem in case of a Continous Distribution Function, in which the r.v. X can assume a continous set of values. In the case of a Discrete Distribution Function the r.v. X can assume a discrete set of values $x_{n}$ with n = 0,1,... , and the P.D.F. in such a case is of the form...

$\displaystyle F(x) = P \{X \le x\} = \sum_{n = 0}^{\infty} P\{X = x_{n}\}\ \mathcal{U} (x - x_{n})\ (3)$

... where $\mathcal{U}\ (*)$ is the Heaviside Step Function. It has to be specified that in this case the derivative of the F(x) in usual meaning doesn't exist...

Kind regards

$\chi$ $\sigma$