i'd like some help with the following problem, please.
Let there be a sequence of random variables $ X_{n}$ such that $n\geq 0$.
If $X_{n} \sim Beta (n,1)$ such that Beta distribution is proportional to $x^{a-1}(1-x)^{b-1}$ for $x \in [0,1]$, find the distribution of $Y_{n} = -nlog(X_{n})$ and an approximation to the probability of the following event:
$\brace \prod^{100}_{n=1}(X_{n})^{n} < e^{-105}$
For the first part I found the following density: \begin{align} f_{Y}&=f_{X}(f^{-1}(Y))\frac{dX}{dY} \\ &=(e^{\frac{-y}{n}})^{a-1}(1-e^{\frac{-y}{n}})^{b-1}e^{\frac{-y}{n}}\frac{-1}{n} \\ &= e^{\frac{-ya}{n}}(1-e^{\frac{-y}{n}})^{b-1}\frac{-1}{n} \end{align} Please correct me if I did something wrong. For the second part, I imagine i must apply logarithm to simplify it and use the distribution abovc.
$B(\alpha,\beta)$ has pdf $ nx^{n-1}$. $$G(t) = \mathbb P(Y_n \leq t) = 1 - \mathbb P(X_n \leq e^{-\frac{t}{n}}) = 1-F(e^{-\frac{t}{n}})$$ thus $$g(t)= \frac{1}{n}f(e^{-\frac{t}{n}})e^{-\frac{t}{n}} = e^{-t}$$ so $Y_n$ is an exponential $1$, i.e. $\Gamma(1,1)$. $$\mathbb P(\Pi_{1}^{100}(X_n)^n < -105) = \mathbb P(\sum_1^{100}Y_n > 105).$$ The sum of indipendent exponential is $\sum_1^{100}Y_n \simeq \Gamma(100,1)$. Thus we need to calculate $$\mathbb P(\Gamma(100,1) > 105)$$ which can be computed with a calculator I guess. Note that its mean is $100$, so $105$ is not that high.