Let N denotes the normally distributed random variable with E[N] = m Var[N] = 0² a) Show the E[L] = exp(E[N] +0.5 Var[N]) = exp (m+ 0.50²) Hint: Start with the probability density function for N: p(x) = (1/(sqrt (2π) σ)) exp( −((x −m)² / (20²))) and use the fact that E[L] = S x p(x) dx b) Obtain the formula for Var[L] in terms of m and σ. Workout E[1/L]. Is it true that E[1/L] = 1/E[L] Let X be a random variable that taken only positive values. Let q(x) denote its pdf. Using integrals, write down the expression equivalent to the condition: E[1/X] =1/E[X] c)

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