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Let $F$ be a distribution function on $\mathbb{R}$ with $F(0)=1$ and $\mu$ be its mean.Show that $$G(x)=\frac{1}{\mu}\int_{0}^{x}[1-F(t)]dt$$ is a distribution function. Also find its mean.

Trial: For $x_1>x_2$ we have $G(x_1)>G(x_2)$. So, it is incrasing.

Again $G(\infty)=\frac{1}{\mu}\int_{0}^{\infty}[1-F(t)]dt=\frac{1}{\mu}\cdot\mu=1$

But how I show $G(-\infty)=\frac{1}{\mu}\int_{0}^{-\infty}[1-F(t)]dt=0$

Lastly help me to find mean. Please help.

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    $\begingroup$ Did you mean to say $F(0)=0$ instead of $F(0) = 1$? The reason for asking is that if $F(0) = 1$, then for $t > 0$, $F(t) = 1$ and so the integrand in the definition of $G(x)$ is $0$ and so $G(x) = 0$ for all $x > 0$. $\endgroup$ – Dilip Sarwate May 7 '13 at 19:37
  • $\begingroup$ @DilipSarwate: I also think so but I check my question and there I find $F(0)=1$. $\endgroup$ – A.D May 7 '13 at 19:46
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If $X$ is a random variable whose cumulative probability distribution function (CDF) $F(\cdot)$ enjoys the property that $F(0) = 0$ (equivalently that $X$ is a nonnegative random variable) then it is well-known that $$E[X] = \int_0^\infty [1-F(t)]\,\mathrm dt.$$ Let us assume that $E[X]$ has finite value $\mu$. Now define the nonnegative function $H(x)$ as $$H(x) = \begin{cases}0, & x \leq 0,\\ \int_0^x [1-F(t)]\,\mathrm dt, & x > 0\end{cases}$$ which increases from $0$ at $x=0$ to a limiting value of $\mu$ as $x \to \infty$. Hence, $G(x) = \mu^{-1}H(x)$ is a CDF since it increases from $0$ at $x=0$ to $1$ as $x \to \infty$.


Note that the definition of $G(x)$ in the question cannot be extended to negative values of $x$ via the integral used; the correct definition is as given above. Also, $F(0) = 1$ which the OP has confirmed is exactly what his textbook says is presumably a typographical error.

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  • $\begingroup$ Nevertheless, assuming $F(0)=0$, $G$ is a valid CDF for all values of $x$, so I think your accusations of inconsistency are a little overblown. There's a simple typographical error in the problem statement, that's all. $\endgroup$ – whuber May 7 '13 at 20:28
  • $\begingroup$ @whuber Thanks for pointing out the mistake. I was mixing up $\mu^{-1}[1-F(t)]$ which is a valid pdf and $G(x)$ which is a valid CDF. I have deleted the incorrect statements. But I do think that for $x < 0$, setting $$G(x)=\frac{1}{\mu}\int_0^x [1-F(t)]dt = -\int_x^0 [1-0]\,\mathrm dt = x$$ as is seemingly implied by the problem is incorrect. $\endgroup$ – Dilip Sarwate May 7 '13 at 20:41
  • $\begingroup$ Thanks. Now I understand how to show that $G(x)$ is C.D.F. But how I find its mean? $\endgroup$ – A.D May 8 '13 at 10:37

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