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Let $F$ be the CDF (=cumulated density function) of a continuous random variable, and $f=F^\prime$ be the PDF (=probability density function). Show that $h$ defined by $$h(x) = \frac{1}{2} \big[\, f(x) + f(-x) \, \big]$$ is also a valid PDF.

My problem: I expected $h(x)$ to have the same support as $f(x)$ -- although this is not stated in the question. However, without assuming that $f(x)$ is an even function, I can only proof this in the limit of an infinit support for $h(x)$.

What I do: I decompose $f(x)$ into an even and an odd part, by defining \begin{align} f(x) &= e(x) + o(x) \\ e(x) &= \frac{f(x) + f(-x)}{2} \\ o(x) &= \frac{f(x) - f(-x)}{2} \end{align} We see, that $h(x)$ is just the even part. Therefore, the odd part $o(x)$ must yield zero, if integrated over the support of $f(x)$. However, I can't proof this.

Assuming that the support of $f(x)$ is given by $x \in [a, b]$, I get \begin{align} 0\stackrel{!}{=} \int_a^b o(x) dx &= ... = \frac{F(x) + F(-x)}{2} \big|_a^b \\ &= \frac{F(b) + F(-b)}{2} - \frac{F(a) + F(-a)}{2} \\ &= \frac{1 + F(-b) - F(-a)}{2} \end{align} So what is left to proof $$F(-b) - F(-a) \stackrel{!}{=} -1$$ As I explain above, for $a\to -\infty$ and $b\to +\infty$ this relationship comes straight from the definition if the CDF.

Any chance we can proof this for a finite support?

  • 1
    You greatly overcomplicate things. To prove that $g(x)$ is a PDF you'd want to check two properties: 1. $g(x)$ is non-negative; 2. $g(x)$ integrates to unity. – 2017-02-26

2 Answers 2

1

Why do you worry about the support so much. For any r.v. its pdf $f$ is $\mathbb{R}\rightarrow\mathbb{R}$ by simply setting $f(x)=0$ for any $x$ the r.v. cannot take. Then $\int_{\mathbb{R}}\frac{1}{2}[f(x)+f(-x)]dx=\frac{1}{2}[1+1]$?

2

In fact $$ h(x)=\frac12[f(x)+f(−x)] $$ is valid $\text{PDF}$ because

  • as $f(x)\ge0$, $h(x)\ge 0$ as well
  • $\int_x h(x)\,dx = \frac12\int_x f(x)\,dx + \frac12\int_x f(-x)\,dx = \frac12+\frac12 = 1$

This conditions are sufficient for $h(x)$ to define valid $\text{PDF}$.