Let X be a random variable exponentially distributed with parameter lambda. The probability density function of X is giv...

Question

Let X be a random variable exponentially distributed with parameter lambda. The probability density function of X is given by:

f_X(x) = lambda * e^{-lambda*x},  x >= 0

If 5 * E[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).

Accepted Answer

lambda = 0.2. For X ~ Exp(lambda), E[X] = 1/lambda and Var(X) = 1/lambda^2. The condition 5*E[X] = Var(X) becomes 5/lambda = 1/lambda^2. Multiplying both sides by lambda^2 gives 5*lambda = 1, so lambda = 1/5 = 0.2.