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Compute the measures of skewness and kurtosis of the Poisson distribution with mean \(\mu\).

Short Answer

Expert verified
The skewness of the Poisson distribution with mean \(\mu\) is \(\frac{1}{\sqrt{\mu}}\) and the kurtosis is \(\frac{1}{\mu}\).

Step by step solution

01

Identify Poisson Distribution

A Poisson distribution is a discrete probability distribution expressing the probability of a given number of events occurring in a fixed interval of time or space. It is characterized by parameter \(\mu\), which is the mean number of successful events.
02

Calculate Skewness

In the case of a Poisson Distribution, skewness, which measures the asymmetry of the distribution, is given by \(\frac{1}{\sqrt{\mu}}\).
03

Calculate Kurtosis

Kurtosis, which measures the 'tailedness' of the distribution, for a Poisson distribution is given by \(\frac{1}{\mu}\).

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Most popular questions from this chapter

Let \(X\) equal the number of independent tosses of a fair coin that are required to observe heads on consecutive tosses. Let \(u_{n}\) equal the \(n\) th Fibonacci number, where \(u_{1}=u_{2}=1\) and \(u_{n}=u_{n-1}+u_{n-2}, n=3,4,5, \ldots\) (a) Show that the pmf of \(X\) is$$ p(x)=\frac{u_{x-1}}{2^{x}}, \quad x=2,3,4, \ldots$$ (b) Use the fact that $$u_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]$$ to show that \(\sum_{x=2}^{\infty} p(x)=1\)

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Let \(F\) have an \(F\) -distribution with parameters \(r_{1}\) and \(r_{2}\). Argue that \(1 / F\) has an \(F\) -distribution with parameters \(r_{2}\) and \(r_{1}\).

Let \(\mathbf{X}=\left(X_{1}, X_{2}, X_{3}\right)\) have a multivariate normal distribution with mean vector 0 and variance-covariance matrix $$\boldsymbol{\Sigma}=\left[\begin{array}{lll}1 & 0 & 0 \\\0 & 2 & 1 \\ 0 & 1 & 2\end{array}\right]$$ Find \(P\left(X_{1}>X_{2}+X_{3}+2\right)\). Hint: Find the vector a so that \(\mathbf{a X}=X_{1}-X_{2}-X_{3}\) and make use of Theorem 3.5.1.

Let \(X\) have the conditional geometric pmf \(\theta(1-\theta)^{x-1}, x=1,2, \ldots\), where \(\theta\) is a value of a random variable having a beta pdf with parameters \(\alpha\) and \(\beta\). Show that the marginal (unconditional) pmf of \(X\) is $$\frac{\Gamma(\alpha+\beta) \Gamma(\alpha+1) \Gamma(\beta+x-1)}{\Gamma(\alpha) \Gamma(\beta) \Gamma(\alpha+\beta+x)}, \quad x=1,2, \ldots$$ If \(\alpha=1\), we obtain $$\frac{\beta}{(\beta+x)(\beta+x-1)}, \quad x=1,2, \ldots$$ which is one form of Zipf's law.

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