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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 \(\displaystyle F(x)\).

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- Thread starter
- #1

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 \(\displaystyle F(x)\).

- Feb 13, 2012

- 1,704

The Poisson distribution is associated with the probability function...

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 \(\displaystyle F(x)\).

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

... 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

$\chi$ $\sigma$

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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 \(\displaystyle F(x)\)

- Feb 13, 2012

- 1,704

The Poisson distribution is aUpto that I have already done. I have confusion about how to find \(\displaystyle F(x)\)

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

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 236

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.The Poisson distribution is adiscretedistribution, 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

$\chi$ $\sigma$

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- Jan 26, 2012

- 4,055

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

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- #8

- Jan 26, 2012

- 4,055

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.

- Feb 13, 2012

- 1,704

In the language of probability F(x) usually indicates theI 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.

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

Its derivative is the

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

All that has no problem in case of a

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

... where $\mathcal{U}\ (*)$ is the

Kind regards

$\chi$ $\sigma$