Density Function of Sum of Exponential Variables

We are given two independent Exponential random variables, X and Y, with means \(\rm \frac{1}{\lambda}\) and \(\rm \frac{1}{\mu}\) respectively, where \(\lambda \ne \mu\). The probability density functions (PDFs) of X and Y are:

• \(f_X(x) = \lambda e^{-\lambda x}\) for \(x > 0\) and \(0\) otherwise.
• \(f_Y(y) = \mu e^{-\mu y}\) for \(y > 0\) and \(0\) otherwise.

We need to find the density function of the sum Z = X + Y. The density function of the sum of two independent random variables is given by the convolution of their individual density functions:

\(f_Z(z) = (f_X * f_Y)(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x) dx\)

Convolution Calculation for Sum of RVs

Since X and Y are non-negative random variables, their densities are non-zero only for positive values. The product \(f_X(x) f_Y(z-x)\) will be non-zero only when \(x > 0\) and \(z-x > 0\), which means \(0 < x < z\). Therefore, for \(z > 0\), the integral limits become:

\(f_Z(z) = \int_{0}^{z} f_X(x) f_Y(z-x) dx\)

Substitute the exponential density functions into the integral:

\(f_Z(z) = \int_{0}^{z} (\lambda e^{-\lambda x}) (\mu e^{-\mu (z-x)}) dx\)

We can pull out the constants \(\lambda\) and \(\mu\):

\(f_Z(z) = \lambda \mu \int_{0}^{z} e^{-\lambda x} e^{-\mu z + \mu x} dx\)

Combine the exponential terms:

\(f_Z(z) = \lambda \mu \int_{0}^{z} e^{-\mu z} e^{(\mu - \lambda) x} dx\)

The term \(e^{-\mu z}\) is constant with respect to the integration variable x, so we can pull it out of the integral:

\(f_Z(z) = \lambda \mu e^{-\mu z} \int_{0}^{z} e^{(\mu - \lambda) x} dx\)

Evaluating the Integral

Now, we evaluate the integral \(\int_{0}^{z} e^{(\mu - \lambda) x} dx\). Since \(\lambda \ne \mu\), \(\mu - \lambda \ne 0\). The integral of \(e^{ax}\) is \(\frac{1}{a}e^{ax}\). Here \(a = \mu - \lambda\):

\(\int_{0}^{z} e^{(\mu - \lambda) x} dx = \left[ \frac{e^{(\mu - \lambda) x}}{\mu - \lambda} \right]_{0}^{z}\)

Evaluate the expression at the limits of integration:

\(= \frac{e^{(\mu - \lambda) z}}{\mu - \lambda} - \frac{e^{(\mu - \lambda) \cdot 0}}{\mu - \lambda}\)

\(= \frac{e^{\mu z - \lambda z}}{\mu - \lambda} - \frac{e^{0}}{\mu - \lambda}\)

\(= \frac{e^{\mu z} e^{-\lambda z} - 1}{\mu - \lambda}\)

Final Density Function

Substitute this result back into the expression for \(f_Z(z)\):

\(f_Z(z) = \lambda \mu e^{-\mu z} \left( \frac{e^{\mu z} e^{-\lambda z} - 1}{\mu - \lambda} \right)\)

Distribute \(\lambda \mu e^{-\mu z}\):

\(f_Z(z) = \frac{\lambda \mu}{\mu - \lambda} (e^{-\mu z} (e^{\mu z} e^{-\lambda z} - 1))\)

\(f_Z(z) = \frac{\lambda \mu}{\mu - \lambda} (e^{-\mu z + \mu z} e^{-\lambda z} - e^{-\mu z})\)

\(f_Z(z) = \frac{\lambda \mu}{\mu - \lambda} (e^{0} e^{-\lambda z} - e^{-\mu z})\)

\(f_Z(z) = \frac{\lambda \mu}{\mu - \lambda} (e^{-\lambda z} - e^{-\mu z})\)

We can rewrite \(\frac{1}{\mu - \lambda}\) as \(\frac{-1}{\lambda - \mu}\). So, the expression becomes:

\(f_Z(z) = \frac{\lambda \mu}{-(\lambda - \mu)} (e^{-\lambda z} - e^{-\mu z})\)

\(f_Z(z) = \frac{\lambda \mu}{\lambda - \mu} (-(e^{-\lambda z} - e^{-\mu z}))\)

\(f_Z(z) = \frac{\lambda \mu}{\lambda - \mu} (e^{-\mu z} - e^{-\lambda z})\)

This density is valid for \(z > 0\). For \(z \le 0\), \(f_Z(z) = 0\).

Comparing this result with the given options, we find that it matches option 3.