Let \(X\) be a random variable exponentially distributed with parameter \(\lambda > 0\). The probability density function of \(X\) is given by:
\[ f_{X}(x) = \begin{cases} \lambda e^{-\lambda x}, & x \ge 0 \\ 0, & \textit{otherwise} \end{cases} \]
If \(5E(X) = Var(X)\), where \(E(X)\) and \(Var(X)\) indicate the expectation and variance of \(X\), respectively, the value of \(\lambda\) is ______ (rounded off to one decimal place).
Answer
0.2
Explanation
For an exponential random variable with rate \(\lambda\):
\[ E(X) = \int_{0}^{\infty} x\lambda e^{-\lambda x}\,dx = \frac{1}{\lambda}, \qquad Var(X) = E(X^{2}) - E(X)^{2} = \frac{2}{\lambda^{2}} - \frac{1}{\lambda^{2}} = \frac{1}{\lambda^{2}} \]
The condition \(5E(X) = Var(X)\) becomes \(\frac{5}{\lambda} = \frac{1}{\lambda^{2}}\). Multiplying by \(\lambda^{2}\), which is allowed because \(\lambda > 0\), gives \(5\lambda = 1\), so \(\lambda = 0.2\).
Taking the mean as \(\lambda\) instead of \(1/\lambda\) is the usual slip; it would give \(\lambda = 5\).
Answer: 0.2