Mathematics Asked on January 1, 2022

For probability triple $(mathbb{R}, mathcal{B}(mathbb{R}), mu)$ prove that for a random variable $X$, if $mu(X>0)>0$, there must be $alpha>0$ s.t. $mu(X>alpha)>0$.

So if $X$ is a random variable with that property, it means that $exists $ event $ A in mathcal{F}$ and interval $B=(0, infty)$ s.t.

$$

A:{omega in mathbb{R}: X^{-1}(B)= A }, mu(A)=m>0

$$

Since $A$ is an interval, we can set its upper and lower bounds as $beta_1, beta_2$. Since $A in mathcal {F}$, we can certainly find number $alpha^{-1}$ such that there exist two disjoint intervals $A_1 cup A_2=A$ with the same measure:

$$

A=A_1 cup A_2, A_1 = [beta_1, alpha^{-1}], A_2 = [alpha^{-1}, beta_2], mu(A_1)=mu(A_2)=frac{m}{2}

$$

Obvisouly $alpha^{-1} in A$, and, since $A$ is a preimage of $B, alpha^{-1} = X^{-1}({alpha})$, and $alpha in B$. Therefore

$$

A_2 = {omega:X^{-1}(alpha, infty)}

$$

and $mu(X>alpha) = mu(A_2)=frac{m}{2}>0$.

I think this is correct, but the hint for the problem is to use the continuity of probabilities, which I didn’t.

The set $E_n={X>frac1n}$ increases to $E={X>0}$. By monotone convergence $0<mu(E)=lim_nmu(E_n)=sup_nmu(E_n)$

From this, it follows that $mu(E_n)>0$ for all sufficiently large $n$.

Answered by Jean L. on January 1, 2022

Consider the expanding sequence of sets $A_n = { X > frac{1}{k}}$. Notice $ bigcup_{k=1}^infty { X > frac{1}{k}} = {X > 0}$. Since $A_n$ is an expanding sequence of sets, then by the Monotone Convergence Theorem for sets $$lim_{ntoinfty} mu(A_n) = muBig(bigcup_{k=1}^infty A_kBig) = mu({X > 0}) > 0$$ We then get that $$0 < muBig(bigcup_{k=1}^infty A_kBig)leq sum_{k=1}^infty mu(A_k)$$ Since the series $sum mu(A_k) > 0$ and each term is nonnegative, then at least one term must be positive, for otherwise $sum mu(A_k) = 0$.

Answered by Andrew Shedlock on January 1, 2022

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