GATE EC-2006

Question

A probability density function density function is of form P(x)= k e ^(-a |x|) , the value of k is

A)0.5

B)1

C) 0.5 a

D) a

Answer 1

Ans C) 0.5a (assuming X $\epsilon$ (-$\infty$, $\infty$))

P(X) = ke^(-a|x|)

$\therefore \int_{-\infty }^{\infty}$ ke^(-a|x|) dx = 1

$\Rightarrow$ 2$\int_{0}^{\infty }$ ke^(-a|x|) dx = 1 ($\because$ ke^(-a|x|) is even function)

$\Rightarrow$ 2k $\int_{0}^{\infty }$ e^-ax dx = 1

$\Rightarrow$ 2k  [ $\frac{e^{-ax}}{-a}$ ]₀^$\infty$ = 1

$\Rightarrow$ $\frac{2k}{-a}$ * (0 - 1) = 1

$\Rightarrow$ k = 0.5a

Comments

thank u..

---

But what happened of e^-a(infinity).  

 

How it become zero as,a is also there