10 Frequentist MLE - M2L3HW1 – Bayesian Statistics

Caution

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--- title: "Frequentist MLE - M2L3HW1" subtitle: "Bayesian Statistics: From Concept to Data Analysis" categories: - Bayesian Statistics keywords: - Distributions· - Homework --- ::::: {.content-visible unless-profile="HC"} ::: {.callout-caution} Section omitted to comply with the Honor Code ::: ::::: ::::: {.content-hidden unless-profile="HC"} [**100 coin flips**]{.column-margin} For Questions (@exr-freq-MLE-1 --- @exr-freq-MLE-3), consider the following scenario: In the example from Lesson 4.1 of flipping a coin 100 times, suppose instead that you observe 47 heads and 53 tails. ::: {#exr-freq-MLE-1} [**100 coin flips**]{.column-margin} Report the value of $\hat p$, the MLE (Maximum Likelihood Estimate) of the probability of obtaining heads. ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: $.47$ This is simply ${47 \over 100}$, the number of successes normalized by the number of trials. ::: ::: {#exr-freq-MLE-2} [**100 coin flips**]{.column-margin} Using the [central limit theorem](/A/A08.qmd) as an approximation, and following the example of [Lesson 4.1](C1-L04.qmd), construct a 95% confidence interval for p, the probability of obtaining heads. Report the lower end of this interval. ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: $.37$ We have $\hat{p} - 1.96\sqrt{\hat{p}(1-\hat{p})/n} =.47 - 1.96\sqrt{(.47)(.53)/100} = .372$, which is the lower end of a 95% confidence interval for p. ::: ::: {#exr-freq-MLE-3} [**100 coin flips**]{.column-margin} Report the upper end of this interval and round your answer to two decimal places. ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: $.57$ We have $\hat{p} + 1.96\sqrt{\hat{p}(1-\hat{p})/n} =.47 + 1.96\sqrt{(.47)(.53)/100} = .568$, which is the upper end of a 95% confidence interval for p. ::: ::: {#exr-freq-MLE-4} [**likelihood function**]{.column-margin} The likelihood function for parameter $\theta$ with data y is based on which of the following? - [ ] $\mathbb{P}r(\theta∣y)$ - [ ] $\mathbb{P}r(y∣\theta)$ - [ ] $\mathbb{P}r(\theta)$ - [ ] $\mathbb{P}r(y)$ - [ ] None of the above. ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: The likelihood is based on the sampling distribution of the data, given the parameter. Note that although the likelihood has the same functional form as $\mathbb{P}r(y\mid\theta)$, it is considered a function of $\theta$. ::: ::: {#exr-freq-MLE-5} [**Mean MLE**]{.column-margin} Recall from Lesson 4.4 that if $X_1,\ldots,X_n \stackrel {IID} \sim Exponential(\lambda)$ (IID means independent and identically distributed), then the MLE for $\lambda$ is $\frac{1}{\bar x}$ where $\bar x$ is the sample mean. Suppose we observe the following data: $X1=2.0, X2=2.5, X3=4.1, X4=1.8, X5=4.0.$ Calculate the MLE for λ. Round your answer to two decimal places. ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: $0.35$ The sample mean is $\bar x = 2.88$. ::: ::: {#exr-freq-MLE-6} [**Exp MLE**]{.column-margin} It turns out that the sample mean $\bar x$ is involved in the MLE calculation for several models. In fact, if the data are independent and identically distributed from a $Bernoulli(p)$, $Poisson(λ)$, or $Normal(μ, σ^2)$, then $\bar x$ is the MLE for $p$, $\lambda$, and $μ$ respectively. Suppose we observe $n=4$ data points from a normal distribution with unknown mean $\mu$. The data are $x=\{−1.2,0.5,0.8,−0.3\}$. What is the MLE for $μ$ ? ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: $-0.05$ This is $(−1.2+0.5+0.8−0.3)/4$. ::: :::::