Understanding the Probability Density Function (PDF) Property

For any valid probability density function (PDF), $f(x)$, the total probability over its entire domain must equal 1. This means the integral of the PDF function from negative infinity to positive infinity must be 1.

Mathematically, this is expressed as: $ \int_{-\infty}^{\infty} f(x) dx = 1 $

In this problem, we are given the PDF: $ f(x) = ae^{-2|x|} $ where $x$ can be any real number, so the domain is $(-\infty, \infty)$.

Calculating the Constant 'a'

We need to find the value of 'a' that satisfies the property $\int_{-\infty}^{\infty} f(x) dx = 1$.

1. Set up the integral equation: $ \int_{-\infty}^{\infty} ae^{-2|x|} dx = 1 $
2. Separate the constant 'a': $ a \int_{-\infty}^{\infty} e^{-2|x|} dx = 1 $
3. Handle the absolute value function: The absolute value function $|x|$ is defined differently for negative and positive values of $x$:
   • $|x| = -x$ if $x < 0$
   • $|x| = x$ if $x \ge 0$
   We need to split the integral into two parts based on this definition: $ \int_{-\infty}^{\infty} e^{-2|x|} dx = \int_{-\infty}^{0} e^{-2(-x)} dx + \int_{0}^{\infty} e^{-2(x)} dx $ $ = \int_{-\infty}^{0} e^{2x} dx + \int_{0}^{\infty} e^{-2x} dx $
4. Evaluate the integrals:
   • First integral: $ \int_{0}^{\infty} e^{-2x} dx = \lim_{b \to \infty} \int_{0}^{b} e^{-2x} dx $ $ = \lim_{b \to \infty} \left[ \frac{e^{-2x}}{-2} \right]_{0}^{b} $ $ = \lim_{b \to \infty} \left( \frac{e^{-2b}}{-2} - \frac{e^{0}}{-2} \right) $ $ = \lim_{b \to \infty} \left( -\frac{1}{2e^{2b}} - \frac{1}{-2} \right) $ As $b \to \infty$, $e^{2b} \to \infty$, so $\frac{1}{2e^{2b}} \to 0$. $ = 0 - \left(-\frac{1}{2}\right) = \frac{1}{2} $
   • Second integral: $ \int_{-\infty}^{0} e^{2x} dx = \lim_{c \to -\infty} \int_{c}^{0} e^{2x} dx $ $ = \lim_{c \to -\infty} \left[ \frac{e^{2x}}{2} \right]_{c}^{0} $ $ = \lim_{c \to -\infty} \left( \frac{e^{0}}{2} - \frac{e^{2c}}{2} \right) $ $ = \frac{1}{2} - \lim_{c \to -\infty} \frac{e^{2c}}{2} $ As $c \to -\infty$, $2c \to -\infty$, so $e^{2c} \to 0$. $ = \frac{1}{2} - 0 = \frac{1}{2} $
5. Combine the results and solve for 'a': $ a \left( \frac{1}{2} + \frac{1}{2} \right) = 1 $ $ a (1) = 1 $ $ a = 1 $

Conclusion

The value of the constant 'a' that makes $f(x) = ae^{-2|x|}$ a valid probability density function is 1.