Masters Degree Level Statistical Question?

Let phiinv(x) be the inverse function of the standard normal distribution function phi(x). Assume that you can efficiently compute both phi(x) and phiinv(x). Let Z ~ N(0,1)





Show that you can generate a random variable Z* from the conditional distrobution of Z given Z %26gt; c by generating random number U ~ U(0,1), letting





Y = U + ( 1 - U )*phi(c)





and setting





Z* = phiinv( Y)





ie.





Prove that P(Z* %26lt;= x) == P( Z %26lt;= x | Z %26gt; c)

Masters Degree Level Statistical Question?
Usually the cdf of the standard normal is Φ(x) (with an uppercase phi) and the pdf is φ(x) (with a lower case phi). I will keep with this notation





Below let f_U(u), f_Y(y), and f_Z*(z*) be the pdf's of U, Y, and Z* respectively.





P(Z ≤ x | Z %26gt; c) = P(Z ≤ x ∩ Z %26gt; c) / P(Z %26gt; c)


= P(c %26lt; Z ≤ x) / P(Z %26gt; c)


= {Φ(x) - Φ(c)} /{1- Φ(c)} if x %26gt; c and 0 otherwise.





So basically, we have to show that


P(Z* ≤ x) = {Φ(x) - Φ(c)} /{1- Φ(c)} if x %26gt; c and 0 otherwise.





let g(u) = u + (1-u)Φ(c), 0 %26lt; u %26lt; 1, which is 1-1 function in u.





Y = U + (1-U)Φ(c)


Y = U + Φ(c) - UΦ(c)


Y - Φ(c) = U(1 - Φ(c))


U = {Y - Φ(c)}/{1 - Φ(c)}





Thus g^-1(y) = {y-Φ(c)}/{1-Φ(c)}, Φ(c) %26lt; y %26lt; 1





f_Y(y) = f_U{g^-1(y)} * |d/dy g^-1(y)|


= 1 * |1/{1-Φ(c)}| (1-Φ(c) is a probability and is always %26gt; 0)


= 1/{1-Φ(c)}, Φ(c) %26lt; y %26lt; 1





This is a constant function in y, which basically means that Y has a uniform distribution with minimum Φ(c) and maximum 1.





let g(y) = Φ^-1(y), Φ(c) %26lt; y %26lt; 1, which is also a 1-1 function in y.





Z* = Φ^-1(Y),


Y = Φ(Z*)





Then g^-1(z*) = Φ(z*), c %26lt; z* %26lt; ∞





f_Z*(z*) = f_Y(g^-1(z*)) * |d/dz* g^-1(z*)|


= 1/{1-Φ(c)} * |φ(z*)| (the normal pdf is always positive)


= φ(z*)/{1-Φ(c)}, c %26lt; z* %26lt; ∞





So P(Z* %26lt;= x) = ∫(c to x) φ(z*)/{1-Φ(c)} dz*


= 1/{1-Φ(c)} * ∫(c to x) φ(z*) dz*


= 1/{1-Φ(c)} * {Φ(z*)}(c to x)


= 1/{1-Φ(c)} * {Φ(x) - Φ(c)}


= {Φ(x) - Φ(c)}/{1 - Φ(c)}





And of course this is only positive if x %26gt; c. Otherwise, the probability is zero.





I was using pdf's pretty much throughout this proof. This problem might be done using cdf's, although I haven't had much experience doing it that way.