# Tag Info

4

There are two steps to solving this problem. The first is based on the realization that what you actually observe are observations $y_i = 1_{(x_i < 0)}$ which are drawn from a Binomial distribution with probability parameter $p(\mu) = \text{P}(x<0 |\, \mu, 1)$, with $\text{P}$ of course being the cumulative Normal distribution. We form the maximum ...

3

$N(x|\mu, \sigma)$ combines the two notations: $x \sim N(\mu, \sigma)$ and $p(x| \mu, \sigma)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma$.

3

The question seems to be asking you to formally state a hypothesis. The "hint" makes a critical fallacy of conflating a "test statistic" with the "effect of interest". In our case we're interested in average price of a phone, it's "expectation". So we either define the average price of either type of phone, or the difference between them. If I were grading ...

2

Here, it means the normal PDF: $$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$ The $\mu,\sigma^2$ in given side means that you can treat them as known quantities.

2

This is a nice and neat simulation that I can demonstrate with R code. I would start off by defining the initial values: population <- 100 consumption_rate <- 5 Next, we could define a few functions to vary the population parameters, as specified in the 3 steps: ## Take a uniform value in 0.01-0.02 as a percentage of our current population ...

1

Indeed, because $X_2+X_3\sim N(0,2)\implies\left(\frac{X_2+X_3}{\sqrt 2}\right)^2\sim \chi^2_1$, independently of $X_1\sim N(0,1)$, $$\frac{\sqrt 2X_1}{|X_2+X_3|}=\frac{X_1}{\sqrt{\left(\frac{X_2+X_3}{\sqrt 2}\right)^2}}\sim t_1$$ Alternatively, since $\frac{X_2+X_3}{\sqrt 2}\sim N(0,1)$, independently of $X_1\sim N(0,1)$, $$\frac{X_1}{\left|\frac{X_2+X_3}{... 1 Begin with the hint given. Take x=0.3. Can you find the value of \theta\in[-1,1] that maximizes \frac{1+0.3\theta}{2}? Take x=0.8. Can you find the value of \theta\in[-1,1] that maximizes \frac{1+0.8\theta}{2}? Take x=-0.5. Can you find the value of \theta\in[-1,1] that maximizes \frac{1-0.5\theta}{2}? Take x=-0.9. Can you find the ... 1 The first step of finding the MLE is to write out the likelihood function (or log-likelihood function). In the present problem, with only a single observed data point, the latter is given by:$$\ell_x(\theta) = \ln(1 + x \theta) + \text{const.} \quad \quad \quad \text{for } -1 \leqslant \theta \leqslant 1. This is the function you need to maximise to ...

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