Almost surely positive.

Question

Hello i have a little question about probability.

I have a function defined on $\mathbb{R}$ as follows,$f(x)=\frac{4x}{(1+x^2)^3}\mathbb{1}_{x>0} $ I have to show that this is the density of an r.v $X$ and that $X>0$ almost surely.

Assuming i've shown that this is its density we have :

$\mathbb{P}(X>0)= \int_{-\infty}^{\infty}\frac{4x}{(1+x^2)^3}\mathbb{1}_{x>0}dx=\int_{0}^{\infty}\frac{4x}{(1+x^2)^3}dx=1$ since this is the density of $X$. Thus X>0 a.s is this correct ?

A bit overkill. Notice that you proved both the positivity a.e. and the fact it is a density — Exodd, Dec 15 '21 at 13:07

score 1 · Accepted Answer · answered Dec 15 '21 at 13:19

Your method is correct but may be unnecessarily long depending on what you mean by "Assuming i've shown that this is its density"

If you've shown it is a density, note $f(x)=0$ for $x\le 0$ so $P(X\le 0)=0$ without any calculation. So $P(X>0)=1$ ie $X$ positive a.s

Almost surely positive.

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