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 ?

  • $\begingroup$ 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). $\endgroup$ – tia Jun 11 at 20:45
  • $\begingroup$ 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 ! $\endgroup$ – 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:

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

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

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