GATE DS&AI 2024 | Question: 47

Question

Let $X$ be a random variable exponentially distributed with parameter $\lambda>0$. The probability density function of $X$ is given by:
\[
f_{X}(x)=\left\{\begin{array}{ll}
\lambda e^{-\lambda x}, \quad x \geq 0 \\
0, & \text { otherwise }
\end{array}\right.
\]

If $5 E(X)=\operatorname{Var}(X)$, where $E(X)$ and $\operatorname{Var}(X)$ indicate the expectation and variance of $X$, respectively, the value of $\lambda$ is $\_\_\_\_\_\_\_\_$ (rounded off to one decimal place).

Answer 1

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 .