Question

4. Let X1, X2, ..., Xn be iid from the Bernoulli distribution with common probability mass function Px(x) = p*(1 – p)1-x for

0 0
Add a comment Improve this question Transcribed image text
Answer #1

p_X(x)=p^x(1-p)^{1-x}

X_1,X_2,....,X_n \textup{ are iid samples.}

\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\therefore \textup{ Likelihood function }L=\prod_{i=1}^n p_{X_i}(x_i)=\prod_{i=1}^np^{x_i}(1-p)^{1-x_i}=p^{\sum_{i=1}^n\!x_i}(1-p)^{n-\sum_{i=1}^n\!x_i}

\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\therefore \textup{ Log-likelihood function}=ln\;L=ln\left(p^{\sum_{i=1}^n\!x_i}(1-p)^{n-\sum_{i=1}^n\!x_i}\right)\\=\left(\sum_{i=1}^n\!x_i\right)\!(ln \;p)+\left(n-\sum_{i=1}^n\!x_i \right )\!(ln(1-p))

\textup{Aim is to maximise log-likelihood function in terms of }p.

\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\frac{\mathrm{d} }{\mathrm{d} p}(ln\;L)=\frac{\mathrm{d} }{\mathrm{d} p}\left[\left(\sum_{i=1}^n\!x_i\right)\!(ln \;p)+\left(n-\sum_{i=1}^n\!x_i \right )\!(ln(1-p))\right]\\ =\frac{\sum_{i=1}^n\!x_i}{p}-\frac{n-\sum_{i=1}^n\!x_i}{1-p}

\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\frac{\mathrm{d} }{\mathrm{d} p}(ln\;L)=0\;\Rightarrow \; \frac{\sum_{i=1}^n\!x_i}{p}-\frac{n-\sum_{i=1}^n\!x_i}{1-p}=0\\ \Rightarrow \;\frac{\sum_{i=1}^n\!x_i}{p}=\frac{n-\sum_{i=1}^n\!x_i}{1-p}\\ \Rightarrow \;\frac{1-p}{p}=\frac{n-\sum_{i=1}^n\!x_i}{\sum_{i=1}^n\!x_i}\\ \Rightarrow \;\frac 1p -1=\frac{n}{\sum_{i=1}^n\!x_i}-1\\ \Rightarrow \;\frac 1p =\frac{n}{\sum_{i=1}^n\!x_i}\\ \Rightarrow \;p=\frac{\sum_{i=1}^nx_i}n=\bar x

\textup{MLE of }p=\hat p_{MLE}=\bar x=\textup{sample mean}

Add a comment
Know the answer?
Add Answer to:
4. Let X1, X2, ..., Xn be iid from the Bernoulli distribution with common probability mass...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT