To find the value of \( \lambda \) such that \( 5E(X) = \text{Var}(X) \), we need to use the properties of the exponential distribution.
For an exponential random variable \( X \) with parameter \( \lambda \), the mean and variance are given by:
\[ E(X) = \frac{1}{\lambda} \]
\[ \text{Var}(X) = \frac{1}{\lambda^2} \]
Given that \( f_X(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \) and \( 0 \) otherwise, we can calculate the expectation and variance as follows:
\[ E(X) = \int_{0}^{\infty} x \cdot \lambda e^{-\lambda x} \, dx \]
\[ = \left[ -x e^{-\lambda x} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-\lambda x} \, dx \]
\[ = 0 + \left[ -\frac{1}{\lambda} e^{-\lambda x} \right]_{0}^{\infty} \]
\[ = \frac{1}{\lambda} \]
Similarly,
\[ E(X^2) = \int_{0}^{\infty} x^2 \cdot \lambda e^{-\lambda x} \, dx \]
\[ = \left[ -x^2 e^{-\lambda x} \right]_{0}^{\infty} + 2\int_{0}^{\infty} x e^{-\lambda x} \, dx \]
\[ = 0 + 2\left[ -x e^{-\lambda x} \right]_{0}^{\infty} + 2\int_{0}^{\infty} e^{-\lambda x} \, dx \]
\[ = 0 + 0 + 2\left[ -\frac{1}{\lambda} e^{-\lambda x} \right]_{0}^{\infty} \]
\[ = \frac{2}{\lambda^2} \]
Now, we can find the variance:
\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]
\[ = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 \]
\[ = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \]
\[ = \frac{1}{\lambda^2} \]
Given that \( 5E(X) = \text{Var}(X) \), we have:
\[ 5 \cdot \frac{1}{\lambda} = \frac{1}{\lambda^2} \]
Solving this equation for \( \lambda \):
\[ 5 \lambda = 1 \]
\[ \lambda = \frac{1}{5} \]
So, the value of \( \lambda \) is \( \frac{1}{5} \) or 0.2 .
