Maximum likelihood estimation (MLE) is a method that can be used to estimate the parameters of a given distribution.
This tutorial explains how to calculate the MLE for the parameter λ of a Poisson distribution.
Step 1: Write the PDF.
First, write the probability density function of the Poisson distribution:
Step 2: Write the likelihood function.
Next, write the likelihood function. This is simply the product of the PDF for the observed values x1, …, xn.
Step 3: Write the natural log likelihood function.
To simplify the calculations, we can write the natural log likelihood function:
Step 4: Calculate the derivative of the natural log likelihood function with respect to λ.
Next, we can calculate the derivative of the natural log likelihood function with respect to the parameter λ:
Step 5: Set the derivative equal to zero and solve for λ.
Lastly, we set the derivative in the previous step equal to zero and simply solve for λ:
Thus, the MLE turns out to be:
This is equivalent to the sample mean of the n observations in the sample.
Additional Resources
An Introduction to the Poisson Distribution
Poisson Distribution Calculator
How to Use the Poisson Distribution in Excel
Awesome! Great help there.
simple and clear, thanks
Thanks for the post! I’m looking for a well-known reference (e.g., book or article) for the log-likelihood function of the Poisson distribution (I need to cite that reference in my paper). I was wondering if you could direct me to that reference. Thanks!
Should step four be d/d\lambda instead of d/dy in the second portion of the equality?
Hi JD…Yes, in step four of the Maximum Likelihood Estimation (MLE) derivation for a Poisson distribution, the differentiation should be with respect to \(\lambda\), not \(y\).
### Step-by-Step Recap of MLE for Poisson Distribution:
We assume that we have a sample \(\{ y_1, y_2, …, y_n \}\) drawn from a Poisson distribution:
\[
P(Y = y \mid \lambda) = \frac{\lambda^y e^{-\lambda}}{y!}, \quad y = 0, 1, 2, \dots
\]
#### 1. **Likelihood Function**
\[
L(\lambda) = \prod_{i=1}^{n} \frac{\lambda^{y_i} e^{-\lambda}}{y_i!}
\]
#### 2. **Log-Likelihood Function**
\[
\log L(\lambda) = \sum_{i=1}^{n} \left( y_i \log \lambda – \lambda – \log y_i! \right)
\]
#### 3. **Differentiate w.r.t. \(\lambda\)**
To find the MLE, we differentiate the log-likelihood with respect to \(\lambda\):
\[
\frac{d}{d\lambda} \log L(\lambda) = \sum_{i=1}^{n} \left( \frac{y_i}{\lambda} – 1 \right)
\]
#### 4. **Set the Derivative to Zero and Solve for \(\lambda\)**
Setting the derivative to zero:
\[
\sum_{i=1}^{n} \left( \frac{y_i}{\lambda} – 1 \right) = 0
\]
\[
\sum_{i=1}^{n} \frac{y_i}{\lambda} = n
\]
\[
\frac{1}{\lambda} \sum_{i=1}^{n} y_i = n
\]
\[
\lambda = \frac{1}{n} \sum_{i=1}^{n} y_i
\]
Thus, the MLE for \(\lambda\) is:
\[
\hat{\lambda} = \frac{1}{n} \sum_{i=1}^{n} y_i
\]
—
### Addressing Your Question:
If in step four, the derivative was mistakenly written as \( \frac{d}{dy} \) instead of \( \frac{d}{d\lambda} \), it would be incorrect because:
– \( y \) is the observed data (fixed), and we are estimating \( \lambda \), so differentiation should be done with respect to \( \lambda \).
– The Poisson likelihood function depends on \( \lambda \), and our goal is to find the MLE of \( \lambda \).
So, yes, in step four, it should definitely be \( \frac{d}{d\lambda} \) and not \( \frac{d}{dy} \).