can some please help me with this question?
sorry i don¨t know how to write in the correct way here.
I know that the gamma and exponential distributions are closely related, if the X_1…X_k are independent and exponentially with parameter lambda, then X_1+….+X_k is gamma distributed with parameters k and lambda. but in my question I have to proof that Y_1 has and exponential distribution with mean 1/alfa, but I can not see any alfa in the fomula of Y_i.
I have already solved a) and b) but I am stuck in c)
Let $X_1,…, X_n, n\in \mathbb{N}$ be independent and identically distributed random variables, and assume that $X_1$ follows a Pareto distribution with a known scale parameter $\theta>0$ and an unknown shape parameter $\alpha>0$. In the following, we examine the estimation of the shape parameter $\alpha$.
a) By differentiating the distribution or survival function, derive that
$g(x;\alpha)=\frac{\alpha\theta^{\alpha}}{(\theta+x)^{\alpha+1}}$
is a probability density function for $X_1$.
The maximum likelihood estimator $\hat{\alpha}$ of $\alpha$ is given by
$\hat{\alpha}=arg max_{\alpha>0} l(\alpha)$
where $l$ is the log-likelihood function:
$l(\alpha)=log\{\mathscr{L}(\alpha)\}, \hspace{2cm} \mathscr{L}(\alpha)=\prod_{i=1}^{n} g(X_i;\alpha)$
b) Show that
$\hat{\alpha}=\frac{n}{\sum_{i=1}^{n}\log \{\frac{\theta+X_i}{\theta}\}}$
For the next two subproblems, we will demonstrate that the maximum likelihood estimator actually exhibits bias, i.e., it satisfies $\mathbb{E}[\hat{\alpha}]\neq \alpha$. For this purpose, let $Y_1,…, Y_n$ be defined as
$Y_i=\log\{\frac{\theta+X_i}{\theta}\}$
c) Show that $Y_1$ is exponentially distributed with a mean of $\frac{1}{\alpha}$, so that $\sum_{i=1}^n Y_i$ is Gamma distributed with shape parameter $n$ and rate parameter$\alpha$.
d) Determine $\mathbb{E}[\hat{\alpha}]$. Propose a new estimator $\hat{\alpha}$ of $\alpha$ that does not exhibit bias, i.e., satisfies $\mathbb{E}[\hat{\alpha}]=\alpha$.
Best Answer
If $Y = h(X)$, where $X$ has pdf $g(x)$ and $h$ is a monotonically increasing or decreasing transformation, we can write the pdf of $Y$ as
$$f(y) = g\left(h^{-1}(y)\right) \cdot \left| \frac{d}{dy} h^{-1}(y) \right |.$$
Here, we have $h(x) = \log\left( \frac{\theta + x}{\theta }\right)$, which means that $h^{-1}(y) = \theta e^y - \theta$. Filling everything in, we get
\begin{align} f(y) &= g\left( \theta e^y - \theta \right) \cdot \left| \frac{d}{dy} ( \theta e^y - \theta ) \right | \\ &= \frac{\alpha \theta^\alpha}{(\theta + \theta e^y - \theta)^{\alpha + 1}}\cdot \left| \theta e^y \right| \\ &= \alpha e^{-\alpha y}. \\ \end{align}