Let the probability density function of a random variable x be given as ƒ(x)=ae^{-2| x |} The value of ‘ a ’ is  ________.

The correct answer is 1.

For a function $f\left(x\right)$ to represent a valid probability density function (PDF) of a continuous random variable, the total area under the probability density curve must equal 1. This normalization condition is mathematically expressed as:

${\int }_{-\infty }^{\infty }f\left(x\right)dx=1$

Substituting the given PDF, $f\left(x\right)=a{e}^{-2\left|x\right|}$, into the normalization equation yields:

${\int }_{-\infty }^{\infty }a{e}^{-2\left|x\right|}dx=1$

The absolute value function $\left|x\right|$ is defined differently for positive and negative values of $x$:
$\left|x\right|=x$ for $x\ge 0$
$\left|x\right|=-x$ for $x<0$

Because the integrand $e^{-2\left|x\right|}$ is symmetric about the y-axis (an even function where $f(-x)=f\left(x\right)$), we can simplify the integral by integrating from 0 to $\infty$ and doubling the result:

$2{\int }_0^{\infty }a{e}^{-2x}dx=1$

Evaluating the integral gives:

$2a{\left[\frac{e^{-2x}}{-2}\right]}_0^{\infty }=1$

Substitute the integration limits:

$2a\left(\underset{x\to \infty }{lim}\frac{e^{-2x}}{-2}-\frac{e^{-2\left(0\right)}}{-2}\right)=1$

Since $e^{-\infty }=0$ and $e^0=1$, this simplifies to:

$2a\left(0-\left(-\frac{1}{2}\right)\right)=1$

$2a\left(\frac{1}{2}\right)=1$

$a=1$

Thus, the value of the constant $a$ is 1.