Math Problem Statement

One common distribution that appears in branching process theory is a DRV with pmf: fX (x; μ) = e−μx(μx)x−1 x! where x ∈ {1, 2, . . .} and μ ∈ (0, 1) (a) Find the MLE for μ given iid X1, . . . , Xn. Then, find the MLE for the particular data x1 = 2, x2 = 1, x3 = 6. (b) Using Desmos, draw a graph of the likelihood function (not log-likelihood) for the data x1 = 2, x2 = 1, x3 = 6. It should be maximal at the μ value you found in part a. Include a sketch of the graph from Desmos (or a screenshot if you’re tech-fancy). (Note: In Desmos, if you click on the wrench icon in the upper-right, you can change the range of values on the x and y axes.)

Solution

Let's address the problem systematically:


(a) Find the MLE for μ\mu

The given pmf is: fX(x;μ)=eμx(μx)x1x!,x{1,2,},μ(0,1).f_X(x; \mu) = e^{-\mu x} \frac{(\mu x)^{x-1}}{x!}, \quad x \in \{1, 2, \ldots\}, \, \mu \in (0, 1).

For a set of iid observations X1,X2,,XnX_1, X_2, \ldots, X_n, the likelihood function is: L(μ;X1,,Xn)=i=1nfX(Xi;μ).L(\mu; X_1, \ldots, X_n) = \prod_{i=1}^n f_X(X_i; \mu).

Substituting fX(x;μ)f_X(x; \mu): L(μ;X1,,Xn)=i=1neμXi(μXi)Xi1Xi!.L(\mu; X_1, \ldots, X_n) = \prod_{i=1}^n e^{-\mu X_i} \frac{(\mu X_i)^{X_i - 1}}{X_i!}.

Simplifying: L(μ;X1,,Xn)=eμi=1nXii=1n(μXi)Xi1Xi!.L(\mu; X_1, \ldots, X_n) = e^{-\mu \sum_{i=1}^n X_i} \prod_{i=1}^n \frac{(\mu X_i)^{X_i - 1}}{X_i!}.

Taking the log-likelihood: (μ)=logL(μ;X1,,Xn)=μi=1nXi+i=1n(Xi1)log(μXi)i=1nlog(Xi!).\ell(\mu) = \log L(\mu; X_1, \ldots, X_n) = -\mu \sum_{i=1}^n X_i + \sum_{i=1}^n (X_i - 1) \log (\mu X_i) - \sum_{i=1}^n \log (X_i!).

Focus on terms involving μ\mu: \ell(\mu) = -\mu \sum_{i=1}^n X_i + \sum_{i=1}^n (X_i - 1) \log \mu + \text{(terms independent of \mu)}.

Differentiating with respect to μ\mu and setting the derivative to zero: μ=i=1nXi+i=1nXi1μ=0.\frac{\partial \ell}{\partial \mu} = -\sum_{i=1}^n X_i + \sum_{i=1}^n \frac{X_i - 1}{\mu} = 0.

Simplify: μi=1nXi=i=1n(Xi1).\mu \sum_{i=1}^n X_i = \sum_{i=1}^n (X_i - 1).

Rearrange: μ=i=1n(Xi1)i=1nXi.\mu = \frac{\sum_{i=1}^n (X_i - 1)}{\sum_{i=1}^n X_i}.


For x1=2,x2=1,x3=6x_1 = 2, x_2 = 1, x_3 = 6:

i=13(Xi1)=(21)+(11)+(61)=1+0+5=6,\sum_{i=1}^3 (X_i - 1) = (2 - 1) + (1 - 1) + (6 - 1) = 1 + 0 + 5 = 6, i=13Xi=2+1+6=9.\sum_{i=1}^3 X_i = 2 + 1 + 6 = 9.

Thus: μ=69=23.\mu = \frac{6}{9} = \frac{2}{3}.


(b) Graph the Likelihood Function

The likelihood function is: L(μ)=eμ9(μ2)1(μ1)0(μ6)52!1!6!.L(\mu) = e^{-\mu \cdot 9} \cdot \frac{(\mu \cdot 2)^{1} \cdot (\mu \cdot 1)^{0} \cdot (\mu \cdot 6)^{5}}{2! \cdot 1! \cdot 6!}.

Simplify constants: L(μ)=Ce9μ(μ)1+0+5,L(\mu) = C \cdot e^{-9\mu} \cdot (\mu)^{1 + 0 + 5}, where CC is independent of μ\mu.

Explicitly: L(μ)μ6e9μ.L(\mu) \propto \mu^6 e^{-9\mu}.


Plot on Desmos

  1. Define L(μ)=μ6e9μL(\mu) = \mu^6 e^{-9\mu}.
  2. Choose a range for μ\mu (e.g., μ(0,1)\mu \in (0, 1)).
  3. Observe the maximum near μ=23\mu = \frac{2}{3}.

Would you like help generating or uploading the graph? I can also generate the graph for you if needed.


Follow-Up Questions:

  1. Why is MLE an important method for parameter estimation in statistics?
  2. What are the practical interpretations of μ\mu in this context?
  3. How does the likelihood function differ from the log-likelihood function?
  4. What are the benefits of using Desmos for visualization in statistical analysis?
  5. What conditions ensure the uniqueness of the MLE in this type of problem?

Tip: Always validate the MLE by checking the second derivative to confirm it is a maximum, not a minimum or inflection point.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Maximum Likelihood Estimation (MLE)
Branching Processes

Formulas

Likelihood function: L(μ; X1, ..., Xn) = Π fX(Xi; μ)
Log-likelihood: ℓ(μ) = -μ Σ Xi + Σ (Xi - 1) log(μ) + constant
MLE for μ: μ = (Σ (Xi - 1)) / (Σ Xi)

Theorems

Properties of Maximum Likelihood Estimation

Suitable Grade Level

Undergraduate Level (Statistics/Probability)