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Verify that the following are probability density functions.$f(x)=\large\frac{1}{\pi}\frac{1}{(1+x^{2})'}$$-\infty$$<$$x$$<\infty $

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Toolbox:
  • The probability density function (continuous probability function $f(x)$ satisfies the following properties :
  • (i) $P(a\leq x\leq b)=\int_a^b f(x) dx$
  • (ii) $f(x)$ is non-negative for all real $x$
  • (iii) $\int_{-\infty}^\infty f(x) dx=1$
  • Also $P(x=a)=0$
  • $P(a\leq x\leq b)=P(a\leq x\leq b)$=P(a < x < b)
Step 1:
We verify whether $f(x)$ satisfies the properties $f(x)\geq 0,-\infty < x <\infty$ and $\int_{-\infty}^\infty f(x) dx=1$
Step 2:
$f(x)=\large\frac{1}{\pi}.\frac{1}{1+x^2}$$ >0$ for all $x$.
Since $1+x^2 >0$ for all $x$
Step 3:
$\int_{-\infty}^\infty f(x) dx=\int_{\infty}^{\infty}\large\frac{1}{\pi}\frac{1}{1+x^2}$$dx$
$\qquad\qquad\;\;=\large\frac{1}{\pi}$$\tan^{-1}x\bigg]_{-\infty}^\infty$
$\qquad\qquad\;\;=\large\frac{1}{\pi}\big[\frac{\pi}{2}-(-\large\frac{\pi}{2})\big]$
$\qquad\qquad\;\;=\large\frac{1}{\pi}$$\pi=1$
Step 4:
From the above,$f(x)$ is a probability density function.
answered Sep 16, 2013 by sreemathi.v
 

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