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There is a statement which I am trying to evaluate to whether true or false:

Let $F_X(x)$ be a cumulative distribution function (cdf) of a r.v. $X$. Then the following $ G_X(x) $ is also a cdf: $$ G(x) = \frac{F_X(x) + 1 - F_X(-x)}{2} $$

It's very clear this is true for distributions which are symmetric around 0, because then $P(X\leq - x) = P(X \geq x)$. It's also simple to show that $ \lim_{x \to \infty} G(x) = 1 $ when $F_X$ is a distribution.

I tend to believe this is a true statement. What is bothering me is the continuity in the right it must have. So I tried to think of a counterexample. Let:

$$ F_X(x) =\begin{cases} 0, \text{ if } x<0 \\ \frac{1}{15} + \frac{2t}{3} \text{ if } x \in [0,1) \\ 1 \text { if } x\geq 1 \end{cases} $$

which is a mixture of a Bernoulli r.v. with $p=4/5$ and an uniform in $(0,1)$ with weights $1/3$ and $2/3$. Now I believe that for $x<0$, $G(x)$ is not continuous in the right (particularly in $x=-1$).

Does it suffice as a counterexample? Or is the statement right?

Thanks!

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You counterexample is correct. And any $F$ that is discontinuous in at least one point will be a counterexample. For example, take $F(x) = \mathbb{I}_{x \geqslant 0}$. Then $G(0) = \frac{1 + 1 - 1}{2} = \frac{1}{2}$, but if $x > 0$ we have $G(x) = 1$, thus $\lim_{x \to 0+} G(x) = 1$.

And in general, $\lim_{x \to x_0+} G(x) = G(x_0) - \lim_{x \to x_0+} \frac{F(-x) - F(-x_0)}{2} = G(x_0) - \lim_{x \to -x_0+}\frac{F(x) - F(-x_0)}{2}$, so $G$ is continuous in right at $x_)$ iff $F$ is continuous in left in $-x_0$.

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  • $\begingroup$ Excelent! Thanks! :) $\endgroup$ – M.Gonzalez Apr 30 '19 at 13:18

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