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Verify that the following are probability density functions.$f(x) = \left\{ \begin{array}{l l} \frac { 2x}{9} & \quad \text{0 $\leq$ x$\leq$3}\\ 0 & \quad \text{elsewhere} \end{array} \right.$

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  • 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:
$f(x)$ is a probability density function if it satisfies the conditions $f(x)\geq 0 -\infty < x< \infty$ and $\int_{-\infty}^\infty f(x)=1$
Step 2:
$f(x)=\large\frac{2x}{9},$$0\leq x\leq 3\Rightarrow f(x)\geq 0$ in this interval (since $x\geq 0)$
$\therefore f(x)\geq 0,-\infty < x< \infty$
Step 3:
Consider $\int_{-\infty}^\infty f(x) dx=\int_0^3\large\frac{2x}{9}$$dx$
(since $f(x)=0$ elsewhere)
$\Rightarrow \large\frac{x^2}{9}\bigg]_0^3$$=1-0=1$
Step 4:
From the above,$f(x)$ is a probability density function.
answered Sep 16, 2013 by sreemathi.v

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