Find the maximum term or terms in the Poisson distribution πk(α), 0 ≤k<∞. Show the same behavior

Find the maximum term or terms in the Poisson distribution...

Key Concepts

Parameter Influence

The parameter ? in the Poisson distribution not only sets the mean and variance but also influences the behavior of the distribution’s terms. Specifically, the value of ? determines whether the maximum term is unique or if the distribution has two adjacent maximum terms, as observed when ? is an integer. This relationship highlights the sensitivity of the distribution's shape and its maximum point(s) to its defining parameter.

Ratio of Successive Terms

Analyzing the ratio of consecutive terms in a probability mass function is a common method for determining the location of the maximum term. In the context of the Poisson distribution, examining the ratio ?_(k+1)(?) / ?_k(?) helps determine the turning point where the probabilities stop increasing and begin decreasing, thereby identifying the mode or modes of the distribution.

Mode of a Discrete Distribution

The mode of a discrete probability distribution is the value at which the probability mass function attains its maximum. For the Poisson distribution, this involves finding the value of k that maximizes ?_k(?). Determining the mode typically requires analyzing the ratio of successive probabilities to understand when the probabilities increase and when they start to decrease.

Poisson Distribution

The Poisson distribution is a discrete probability distribution that models the number of events occurring in a fixed interval of time or space, given that these events happen with a known constant mean rate and independently of each other. Its probability mass function is characterized by the parameter ?, which serves both as the mean and the variance of the distribution.

Transcript

00:01 Here, it is given that a random variable x follows poised distribution with parameter lambda.

00:09 So, expectation of x, expectation of x is also lambda.

00:16 Now, it is based on the previous problem where lambda is an integer.

00:24 So lambda can only take integer values.

00:27 And it is given that it is 2, 4, 6, up to 18.

00:35 Now, if parameter of a poised distribution is an integer, then the distribution is bi -modal.

00:41 So, what is the meaning of by -modal? that means the distribution has two modes, and those are lambda and lambda -minus.

00:50 That means the pmf -px peaks at two values at x equals to lambda and lambda -minus 1.

01:04 Now, using two inequalities, those are px plus 1, whole divided by px is greater than 1, another is px plus 1, whole divided by px is less than 1.

01:23 Using these two inequalities, we have to show that the pmf picks at x equals to lambda and x equals to lambda minus 1.

01:32 Okay, now let's start with the problem.

01:37 So first we write px plus 1 whole divided by px is greater than 1.

01:47 First inability.

01:48 Now the probability mass function of this perjure distribution is px equals to e to the power minus lambda, lambda to the bar x, whole divided by x factorial.

01:59 Where x equals to 0 1 2 and so on and lambda greater than 0 and 0 otherwise okay so these in inequality if we write e to the power minus lambda lambda to the power x plus 1 whole divided by x plus 1 factorial into x factorial whole divided by e to the bar minus lambda lambda lambda to the bar x this quantity a greater than 1.

02:32 So this equals to lambda whole divided by x plus 1 is greater than 1.

02:39 That implies x plus 1 is less than lambda.

02:45 So x less than lambda minus 1.

02:48 So that is see as a as lambda takes only integer values...

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