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A continuous random variable $x$ has the p.d.f defined by $f(x) = \left\{ \begin{array}{1 1} ce^{-ax} & \quad 0 < x < \infty \\ 0 & \quad \text{else where} \end{array} \right.$ Find the value of $c$ if $a>0$

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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:
The probability density function of a continuous random variable is
$f(x)=\left\{\begin{array}{1 1}ce^{-ax},&0 < x < \infty\\0,&else\;where\end{array}\right.$ where $a > 0$
By definition of a probability density function $\int_{-\infty}^\infty f(x) dx=1$
Step 2:
$\int_{-\infty}^\infty f(x) dx=1$
$\Rightarrow \int_0^{\infty} ce^{-ax}dx=1$
(Since $f(x)=0$ elsewhere)
$\therefore \large\frac{ce^{-ax}}{-a}\bigg]_0^\infty$$=1$
$\Rightarrow [0+\large\frac{c}{a}]$$=1$
$\Rightarrow c=a$
answered Sep 17, 2013 by sreemathi.v

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