Question

Let X be a random number from (0,1). Find the probability density functions of the random variables $Y = -ln(1-X) and Z =X^{n}, n\neq 0$

Answer 1

be random number from . So it can be considered as uniform. We can write the density as

$f_X(x)=1\quad 0<x<1$.

Let $Y=-\ln (1-X)$

$Y=-\ln (1-X)\\ \Rightarrow -Y=\ln(1-X)\\ \Rightarrow e^{-Y}=1-X\\ or \quad X=1-e^{-Y}$

When x=0, y=0 and when $x=1 , y=\infty$

$\frac{dx}{dy}=e^{-y}$

So the density of Y is given by

$f_Y(y)=f_X(x)\left|\frac{dx}{dy}\right|\\=1.e^{-y}\\ =e^{-y},\quad 0<y<\infty$which is the exponential density.

So $Y=-\ln(1-X)$ has exponential distribution with probability density function

$f_Y(y)=e^{-y},\quad 0<y<\infty$.

Now we want to find the density of $Z=X^n$

Whenx=0,z=0 and when x=1, z=1.

$Z=X^n\Rightarrow X=Z^{\frac{1}{n}}$

$\frac{dx}{dz}=\frac{1}{n}z^{\frac{1}{n}-1}=\frac{1}{n}z^{\frac{1-n}{n}}$

So the density of Z is given by

$f_Z(z)=f_X(x)\left|\frac{dx}{dz}\right|\\ =1.\frac{1}{n}z^{\frac{1-n}{n}}\\ =\frac{1}{n}z^{\frac{1-n}{n}},\quad 0<z<1$

So the probability density function of Z is given by $f_Z(z)=\frac{1}{n}z^{\frac{1-n}{n}},\quad 0<z<1$ .