Calculating distribution function $Y = \min{X_1, ..., X_n}$ and calculating the density of $Y = -\log(X_1)$

Question

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!

Answer 1

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 $Y\sim \exp(1)$

Answer 2

$X_1,...,X_n \sim U(0,1)$, $Y=min(X_1,...,X_n)$. We can consider finding the cdf of $Y$ and differentiate to find the pdf of $Y$.

Consider $$P(Y \leq y)=P(min(X_1,...,X_n) \leq y)=1-P(min(X_1,...,X_n) \geq y)$$ $$=1-P(min(X_1,...,X_n) \geq y)=1-P(X_1 \geq y \cap ... \cap X_n \geq y)$$ From the independence of $X_1,..,X_n$, $$=1-P(X_1 \geq y)...P(X_n \geq y)=1-(\int_{y}^{1}1dx)...(\int_{y}^{1}1dx)$$ $$F_y(y)=1-(1-y)^{n}$$ $$f_Y(y)=n(1-y)^{n-1}, \text{for y $\in (0,1)$}$$

For the second question of $Y=-\log(X_1)$, we can still use the method of cdf. $$P(Y \leq y)=P(-log(X_1)\leq y)=P(X_1 \geq e^{-y})=\int_{e^{-y}}^{1}1dx$$ $$F_Y(y)=1-e^{-y}$$ So, $$f_Y(y)=e^{-y}, \text{for y $\in (0,\infty)$}$$

Hopefully I did not make a mistake, if I did, please comment down below. Hope this helps :)