is strictly increasing at values of which are possible for the inverse F x* X, function is welldefined. Then P[ ] = P[ ] = P[ 1(u) ] = 1(u) ) = U u F(X) u X F F( F u ≤ ≤ ≤ Done. Here’s a side note on finding inverse functions. If the inverse is y = g(x), g1 found just by solving for in the equation (if possible). The result will x y = g(x) give If = + 1 the solution is = = . Both and are handled as g1. y x3 x x y symbols, so you’ll see statements like these: “dummy” The inverse function of + 1 is . x3 The inverse function of is . x3 Care is needed for many-to-one functions. The function = maps both = 2 y x2 x and = +2 to the same value of so the square function does not have an inverse. x y, If we can agree that we’re only dealing with positive values of then the inverse x, of 2 is the positive square root . x If we are considering sin where ranges over all real numbers, then sin x, x x does not have an inverse. If we can agree that the only of concern are in the x’s inverval (, +), then sin1 can be defined. It helps to see the theorem in the context of the example. For that example, = . F(X) Let’s ask then . = = The here is base e. “log” = = Since is a value in (0, 1), it must happen that 1 is u u – also in (0, 1). So log(1 is negative. u) – = = = 1 (1 = u) u – Here’s how this is used. Let’s suppose that you want to simulate values X1, X2, X3, … from a continuous cumulative distribution function Create the random uniform F.
F is strictly increasing at values of x* which are possible for X, the inverse function is welldefined. Then P[ U ≤ u ] = P[ F(X) ≤ u ] = P[ X ≤ F 1(u) ] = F( F 1(u) ) = u Done. Here’s a side note on finding inverse functions. If y = g(x), the inverse g1 is found just by solving for x in the equation y = g(x) (if possible). The result will give g1. If y = x3 + 1 the solution is x = = . Both x and y are handled as “dummy” symbols, so you’ll see statements like these: The inverse function of x3 + 1 is . The inverse function of x3 is . Care is needed for many-to-one functions. The function y = x2 maps both x = 2 and x = +2 to the same value of y, so the square function does not have an inverse. If we can agree that we’re only dealing with positive values of x, then the inverse of x 2 is the positive square root . If we are considering sin x, where x ranges over all real numbers, then sin x does not have an inverse. If we can agree that the only x’s of concern are in the inverval (, +), then sin1 can be defined. It helps to see the theorem in the context of the example. For that example, F(X) = . Let’s ask then . = = The “log” here is base e. = = Since u is a value in (0, 1), it must happen that 1 – u is also in (0, 1). So log(1 – u) is negative. = = = 1 (1 – u) = u Here’s how this is used. Let’s suppose that you want to simulate values X1, X2, X3, … from a continuous cumulative distribution function F. Create the random uniform
SimulationTheorem.docx