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)
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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Please provide explanantion
![Let N denotes the normally distributed random variable with
E[N] = m
Var[N] =²
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 (2µ) σ)) exp( −((x − m)² / (20²))) and use the fact that
E[L] = S x p(x) dx
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]
b)
c)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00828718-2df8-4d22-87de-a2e53e10f924%2F383a1f39-7667-4044-ab93-794098fa98ec%2Fafcffck_processed.png&w=3840&q=75)
Transcribed Image Text:Let N denotes the normally distributed random variable with
E[N] = m
Var[N] =²
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 (2µ) σ)) exp( −((x − m)² / (20²))) and use the fact that
E[L] = S x p(x) dx
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]
b)
c)
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