# Question 1 (Substitute actual equations for pk/pc): ### Given: The number of packs of toilet paper in Kroger follows a distribution with parameter $\lambda$: $$p_K(x)=\frac{2^x\lambda^xe^{-2\lambda}}{x!}$$ And the number of packs of toilet paper in Costco follows a distribution with parameter $\mu$: $$p_C(x)=\frac{\mu^xe^{-\mu/2}}{2^xx!}$$ | (Observance $i$) | Store z^($i$) | Storage x^($i$) | |--------------|-----------------------|-------------------------| |--------------|-----------------------|-------------------------| | 1 | Kroger | 5 | | 2 | Kroger | 8 | | 3 | Costco | 4 | | 4 | Kroger | 2 | | 5 | Costco | 5 | | 6 | Costco | 10 | | 7 | Kroger | 10 | | 8 | Costco | 2 | | 9 | Costco | 2 | | 10 | Costco | 3 |

a. Write out the likelihood of this dataset, substituting in the probability distributions of $p_K(x)$ and $p_C(x)$. Leave your answer as a product of terms, without substituting in any numerical values. Answer: The likelihood of this dataset can be also written as the intersection of the probabilities of each independent event. Simply, take A, which is the likelihood of observation 1, and probabilities B-J for probabilities 2-10, and intersecting them. We can do this, as each and every observation is an independent event and has no effect on other observations. Therefore, we simply do > $$\prod\limits_{i=1, i\in{\{1,2,4,7\}}}^{10} \gamma _K \frac{2^{x^{(i)}}\lambda^{x^{(i)}}e^{-2\lambda}}{x^{(i)}!} * \prod\limits_{i=1, i\in{\{3,5,6,8,9,10\}}}^{10}\gamma _C \frac{\mu^{x^{(i)}}e^{-\mu/2}}{2^{x^{(i)}}x^{(i)}!}$$ b. Recall that with MLE, we want to find the parameters that maximize the dataset’s likelihood. Sometimes, it is easier to maximize the log-likelihood rather than directly maximizing the likelihood. Using your answer from part (a), find the log-likelihood for this dataset. Represent your final answer in terms of [insert variables here] Answer: $$ln(\prod\limits_{i=1, i\in{\{1,2,4,7\}}}^{10} \gamma _K \frac{2^{x^{(i)}}\lambda^{x^{(i)}}e^{-2\lambda}}{x^{(i)}!}) + ln(\prod\limits_{i=1, i\in{\{3,5,6,8,9,10\}}}^{10}\gamma _C \frac{\mu^{x^{(i)}}e^{-\mu/2}}{2^{x^{(i)}}x^{(i)}!})$$
$$= \Sigma_{i=1, i\in{\{1,2,4,7\}}}^{10} ln(\gamma _K \frac{2^{x^{(i)}}\lambda^{x^{(i)}}e^{-2\lambda}}{x^{(i)}!}) + \Sigma_{i=1, i\in{\{3,5,6,8,9,10\}}}^{10}ln(\gamma _C \frac{\mu^{x^{(i)}}e^{-\mu/2}}{2^{x^{(i)}}x^{(i)}!})$$
$$= \Sigma_{i=1, i\in{\{1,2,4,7\}}}^{10} ln(\gamma _K) + ln(2^{x^{(i)}}) + ln(\lambda^{x^{(i)}}) + ln(e^{-2\lambda}) -ln(x^{(i)}!) + \Sigma_{i=1, i\in{\{3,5,6,8,9,10\}}}^{10}ln(\gamma _C) + ln(\mu^{x^{(i)}}) + ln(e^{-\mu/2}) - ln(2^{x^{(i)}}) - ln(x^{(i)}!)$$
$$= \Sigma_{i=1, i\in{\{1,2,4,7\}}}^{10} \ln(\gamma _K) + x^{(i)}\ln(2) + x^{(i)}\ln(\lambda) -2\lambda - ln(x^{(i)}!) + \Sigma_{i=1, i\in{\{3,5,6,8,9,10\}}}^{10}\ln(\gamma _C) + x^{(i)}\ln(\mu) + \frac{-\mu}{2} - x^{(i)}\ln(2) - \ln(x^{(i)}!)$$
$$= -S_K-S_C+(\ln(2)+\ln(\lambda)\cdot m_K+(\ln(\mu)-\ln(2))\cdot m_C-8\lambda -3\mu + 4\ln(\gamma_K)+6\ln(\gamma_C)$$ c. Calculate the maximum likelihood estimation for γc, γk, λ, and µ by maximizing the expression for log-likelihood that you derived in part (b). Express your final answer as numbers, and be sure to show all your work. [Hint: Remember that γC + γK = 1] Partial derivatives are trivial to compute, so work will be minimal Partial with respect to $\gamma_K$: $\frac{4}{\gamma_K}$-$\frac{6}{1-\gamma_K}$ Partial with respect to $\gamma_C$: 1-$\frac{2}{5}$ Partial with respect to $\lambda$: -8+$\frac{m_K}{\lambda}$ Partial with respect to $\mu$: -3+$\frac{m_C}{\mu}$ As defined $m_K=25$, and $m_C=26$ Therefore $\mu=\frac{m_C}{3}$ and $\lambda=\frac{m_K}{8}$. This means $\mu=\frac{26}{3}$ and $\lambda=\frac{25}{8}$