Show that E[e^{sigma Z} 1{Z > a}] equals e^{sigma^2/2} N(sigma - a) for Z standard normal, and evaluate it at sigma = 0.2 and a = 0.5.

Show that E[e^{sigma Z} 1{Z > a}] equals e^{sigma^2/2} N(sigma - a) for Z standard normal, and evaluate it at sigma = 0.2 and a = 0.5.

Approach: Write the expectation as an integral of e^{sigma z} against the standard normal density, complete the square in the exponent, and read the shifted region of integration as a normal distribution function.

0.3898. Write the expectation as the integral over z > a of e^{sigma z} e^{-z^2/2}/sqrt(2 pi) dz. Completing the square gives sigma z - z^2/2 = -(z - sigma)^2/2 + sigma^2/2, so the expression is e^{sigma^2/2} times the integral over z > a of the normal density evaluated at z - sigma. Substituting y = z - sigma turns the region into y > a - sigma, and by symmetry of the density that probability is N(sigma - a). Numerically e^{0.02} = 1.0202 and N(0.2 - 0.5) = N(-0.3) = 0.3821, so the value is 0.3898. This is the whole content of the first term of the Black-Scholes formula: the factor e^{sigma^2/2} is the drift correction and the shift from a to a - sigma is the measure change to the stock numeraire, which is why the two normal arguments d_1 and d_2 differ by exactly sigma sqrt(T).

Follow-up: Using this identity, write the price of a digital call and of an asset-or-nothing call, and say which one a trader hedges more easily near expiry?

Key concepts: completing the square, normal density, measure change, black-scholes formula.