lebesgue distribution in probability

Let's assume $$\Omega=[0,1]$$ and $$F$$ is a Borel set of $$[0,1]$$, and $$\mathbb{P}$$ is Lebesgue's measure in $$[0,1]$$. Also let's assume that $$\forall n \geq 1$$, $$X_n$$ is a random variable defined such $$X_n(\omega)=\omega^n$$ and of course $$\omega \in \Omega$$.

How do I define the density of $$X_n$$ ?

I've tried to work on its repartition function like this :

$$\mathbb{P}(X_n \leq t)= \int_{[0,1]} ω^n.\mathbb{1}_{\omega \leq t} d\omega$$

then discussing both cases when $$t \in [0,1]$$ or not, but I seem to find zero which is weird since it should be something that depends on t that I would after that derive to finally find my density function.

What seems to be wrong in my work ? can you help me find my error please ?

• Welcome to MSE. Just a small note regarding your writing. When you write a mathematical expression, for example t \in [0,1], you should enclose the entire expression with dollar signs, and not just \in (as you tended to do on this post, which I edited). Note the difference: $t \in [0,1]$ (correct), t $\in$ [0,1] (incorrect). – tia Jun 11 at 20:45
• I wouldn't do that on purpose, please understand that i'm no familiar with mathjax nor lateX and I was very meticulous trying to write everything as correctly as I could, nonetheless thanks for the advice ! – Blueberry Jun 11 at 20:55

The probability that $$X_n(\omega)=\omega^n\leq x$$ is the Lebesgue measure of the set

$$\{\omega \in [0,1] | \omega^n\leq x \}=\{\omega\in [0,1] |\omega\leq x^{\frac 1 n}\}$$ which is

$$x^{\frac 1 n}, x\in [0,1],$$

and zero below $$0$$, and $$1$$ above $$1$$.

The derivative of the function above is the density of $$X_n$$.

Your mistakes are shown in red below:

$$\mathbb{P}(X_n \leq t)= \int_{[0,1]} \color{red}{ω^n}.\mathbb{1}_{\omega^{\color{red}n} \leq t} d\omega$$

• Oh my, what a stupid mistake, thank you very much – Blueberry Jun 13 at 10:25