Let X be a Poisson distributed random variable such that P[X=0]=0.5. Find the mean of X.

Name: Problem 24, Chapter 2: Fundamentals of Mathematical Statistics
Uploaded: 2023-06-27T16:58:04-08:00
Duration: 7 min 10 s
Channel: Aparna Shakti
Description: Let X be a Poisson distributed random variable such that P[X=0]=0.5. Find the mean of X.

Let X be a Poisson distributed random variable such that...

Key Concepts

Poisson Distribution

The Poisson distribution is a discrete probability distribution that models the number of events occurring within a fixed interval of time or space under the assumption that these events occur independently and at a constant average rate. Its key parameter, ? (lambda), represents both the mean and the variance of the distribution.

Probability Mass Function and Exponential Relationship

The probability mass function (PMF) for a Poisson random variable is given by P(X = k) = (e^(??) ?^k) / k! for nonnegative integers k. In particular, the probability of observing zero events (k = 0) is P(X = 0) = e^(??), and this relationship can be used to determine ? when the probability of zero events is provided.

Transcript

00:01 Let x be a poisson distributed random variable such that p of x is equal to 0 and this value is basically equals to 0 .5.

00:13 So find the mean of x.

00:15 This is required.

00:16 So in poisson distribution mean that is mu is equal to the parameter lambda.

00:25 So which represents the average number of events or occurrences in a given interval.

00:33 So here mu when you talk about it is equals to parameter lambda itself.

00:39 So now what is given to us from here we are going to calculate the value of mean here.

00:52 So a mean of x we need to find out.

00:54 So this is given to us p of x is equals to 0 and this value is basically equals to 0 .5.

01:00 So this is the probability right at x is equals to 0.

01:04 So now probability mass function that is pamf of poisson distribution representation is done by which expression by this expression.

01:13 What is the expression? p is equal to p of x is equals to k is equals to e to the power minus lambda multiplied by lambda to the power k.

01:31 So this is basically the formula upon factorial k right.

01:35 So this is the formula right.

01:37 So from here you can conclude that from equation 1 putting the value values of equation 1 in equation 1 in above equation above equation right we have what we have p of x is equals to 0 here right.

02:13 So that is equals to 0 .5 itself which is also equals to e to the power minus lambda right and lambda to the power 0 because the value of k is 0 here right 0 this value is 0.

02:31 Now factorial k that is factorial 0.

02:35 So factorial 0 is nothing but that is equals to 1 itself right.

02:40 So now we are getting on solving we will be getting e to the power minus lambda is equals to 0 .5.

02:47 We will compare these two equations right.

02:50 So on comparing what you could say taking log on both sides both sides.

03:00 So basically when you talk about the log so you take log that is ln.

03:06 Ln means log with base n.

03:09 So ln e to the power like e to the power minus lambda right is equals to ln 0 .5...