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Chapter 3: Problem 11

Let \(X\) have a Poisson distribution. If \(P(X=1)=P(X=3)\), find the mode of the distribution.

Short Answer

Expert verified
The mode of the distribution is \( \sqrt{6} - 1 \).

Step by step solution

01

Initializing the exercise

We are given that X has a Poisson distribution. In a Poisson distribution, the probability of an event occurrence is given as: \[ P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where \(\lambda\) is both the mean and variance of the distribution, e is the base of the natural logarithm, and k is the number of occurrences of an event (which can be any non-negative integer). We are given that \(P(X=1)\) is equal to \(P(X=3)\).
02

Setting up the equations

We substitute the given event occurrences into the formula for the Poisson distribution to form two equations. \[ P(X=1) = \frac{\lambda^1e^{-\lambda}}{1!} = e^{-\lambda} \lambda \] \[ P(X=3) = \frac{\lambda^3e^{-\lambda}}{3!} = e^{-\lambda} \frac{\lambda^3}{6}\] Since P(X=1) = P(X=3), we can set the two expressions equal to each other and solve for \(\lambda\).
03

Solving the equation

Setting e^{-\lambda} \lambda = e^{-\lambda} \frac{\lambda^3}{6} and solving for \(\lambda\) gives the cubic equation \(\lambda^3 - 6\lambda = 0\). This equation can be factored to \(\lambda(\lambda - \sqrt{6})(\lambda + \sqrt{6}) = 0.\) Since \(\lambda\) cannot be negative, the solutions to the equation are \(\lambda = 0 \) and \(\lambda = \sqrt{6}\). However, since the probabilities of events occurring is non-zero, the only feasible solution is \(\lambda = \sqrt{6}\).
04

Finding the mode

Since \( \lambda \) is not an integer, the mode of the Poisson distribution is \( \lambda - 1 \). Therefore, the mode of the distribution is \( \sqrt{6} - 1 \).

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Most popular questions from this chapter

Let \(X\) and \(Y\) have the joint pmf \(p(x, y)=e^{-2} /[x !(y-x) !], y=0,1,2, \ldots\), \(x=0,1, \ldots, y\), zero elsewhere. (a) Find the mgf \(M\left(t_{1}, t_{2}\right)\) of this joint distribution. (b) Compute the means, the variances, and the correlation coefficient of \(X\) and \(Y\). (c) Determine the conditional mean \(E(X \mid y)\). Hint: Note that $$\sum_{x=0}^{y}\left[\exp \left(t_{1} x\right)\right] y ! /[x !(y-x) !]=\left[1+\exp \left(t_{1}\right)\right]^{y}$$ Why?

Let \(X, Y\), and \(Z\) have the joint pdf $$\left(\frac{1}{2 \pi}\right)^{3 / 2} \exp \left(-\frac{x^{2}+y^{2}+z^{2}}{2}\right)\left[1+x y z \exp \left(-\frac{x^{2}+y^{2}+z^{2}}{2}\right)\right]$$ where \(-\infty

Let \(X\) be a random variable such that \(E\left(X^{m}\right)=(m+1) ! 2^{m}, m=1,2,3, \ldots\). Determine the mgf and the distribution of \(X\).

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