Let $X_1,..,X_n \sim U(0,1)$ be independent and uniformly distributed on the interval $[0,1]$.
How can one
- calculate the distribution function of the stochastic variable $Y = \min{X_1, …, X_n}$
- calculate the density of $Y = -\log(X_1)$
Regarding the first question, I have
$$F_{Y}(x) = P(Y \leq x) = 1- (1-F(x))^2 $$
Regarding the second question, I define
$Y_1 = -\log(X_1)$ and $Y_2 = -\log(X_2)$. Since $Y_1$ and $Y_2$ are independent and uniformly distributed, we have
\begin{align}
P(Y_1 / (Y_1 + Y_2) \leq t) &= P(Y_1 + Y_2 \geq >_1 / t) \\
&= P(Y_2 \geq Y_1 (1 / t – 1)) \\
&= \text{E} \left [ P(Y_2 \geq Y_1 (1 / t – 1) \mid Y_1) \right ] \\
&= \text{E} \left ( e^{-Y_1 (1 / t – 1)} \right ) \\
&= \int_{0}^{\infty} e^{-s (1 / t – 1)} e^{-s} ds \\
&= \int_{0}^{\infty} e^{-s / t} ds \\
&= t
\end{align}
which represents the uniform distribution function.
Is that correct/wrong?
Any help is appreciated!
Best Answer
Regarding the first answer,
$$F_Y(y)=1-(1-F_X(y))^n=1-(1-y)^n$$
Regarding the second question,
$$y=-\log x_1$$
$$x_1=e^{-y}$$
$$|x_1'|=e^{-y}$$
Thus
$$f_Y(y)=e^{-y}$$
That means $Y\sim \exp(1)$