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Start with $X\sim N(\mu;\sigma^2)$. This means that X is normally distributed with mean $\mu$ and standard deviation $\sigma$ If you transform you rv in the following way $$Z=\frac{X-\mu}{\sigma}$$ yo …

The quick answer is that $Y=e^X\sim \text{Lognormal}$ thus its mean is well known If you want to do all the calculation with the gaussian distribution, it is not difficult; try, it is a good exercise

I did not do all the calculations because it is only a matter to solve algebraic systems but I explain you how to do... To calculate MoM's estimators, the first thing you have to do is to express your …

First question: if Bob's probability is the half of the other 49 this means that $$49\times 2p+p=1$$ $$p=\frac{1}{99}$$ Now you can proceed with the second question

Your expression is also a measure of dispersion. In fact it is a distance between the data points and $\mu$. It is not optimal becasuse (it is easy to prove) that the minimum distance $$\int_{-\infty} …

Now let's suppose we have to calculate the power of this test. Starting from the Decision rule that is to reject $H_0$ if $Y=\sum_i X_i>21.3$ the power is defined as follows $$\gamma=\mathbb{P}[Y>21.3 …

there is a minor error in the first integral too... $$f_{YZ}(y,z)=\int_{-\infty}^{\infty}\left( \frac{1}{\sqrt{2\pi}} \right)^3e^{-[x^2+(y-x)^2+(z-x)^2]/2}dx$$ that is $$f_{YZ}(y,z)=\frac{1}{2\pi}e^{- …

Your density is the one of a Pareto distribution. In the enclosed link you will find your result which is easy to get simply integrating $f_X$.

There are no issues in this exercise. The absolute value, related to the data, does not affect $\theta$ estimation Your likelihood is $$L(\theta)\propto \theta^n\prod_i|x_i|^\theta$$ $$l(\theta)=n\log …

When you propose a Gamma density you must give information in order to understand which parametrization you are using. Given that you assume $E(X)=\alpha\cdot\beta$ your density is the following $$f_X …

Ok Let's start! $$p(\theta|\mathbf{y}) \propto Exp\Bigg\{-\frac{\theta^2}{2\sigma_0^2}\Bigg\}Exp\Bigg\{-\frac{1}{2}\Sigma_i(y_i-\theta x_i^2)^2\Bigg\}$$ Now let's expand the exponent throwing away any …

Example $X_1,\dots X_5$ is a random sample drawn from a population $X\sim U(0;\theta)$ These are statistics: $T_1=\Sigma_i X_i$ $T_2=\frac{1}{5}\Sigma_i X_i$ $T_3=\max(X_i)$ $T_4=\min(X_i)$ $T_5=\ …

Hint for the solution First define the Likelihood fuction, that is $$L(\alpha;\lambda)=\lambda^ne^{-\lambda \sum_i x_i}e^{n \alpha \lambda}\mathbb{1}_{(-\infty; x_{(1)}]}(\alpha)$$ Find the MLE est …

First observe that $$\int_k^{\infty}x \phi(x)dx$$ is not your $E(Y)$ (but sure you already know this) Example: $X\sim N(0;1)$ and $k=4$ Without any calculation, $E(Y)\approx 4$ because the probabilit …

does the unbiased estimator qualify as being one if it is a function of the parameter it is trying to estimate? an estimator, biased or not, cannot be a function of the parameter....an estimator, by …