Transformations of Ranom Variables September, 2009 We begin with a ranom variable an we want to start looking at the ranom variable Y = g() = g where the function g : R R. The inverse image of a set A, In other wors, g (A) = { R; g() A}. g (A) if an only if g() A. For eample, if g() = 3, then g ([, 8]) = [, 2] For the singleton set A = {y}, we sometimes write g ({y}) = g (y). For y = 0 an g() = sin, g (0) = {kπ; k Z}. If g is a one-to-one function, then the inverse image of a singleton set is itself a singleton set. In this case, the inverse image naturally efines an inverse function. For g() = 3, this inverse function is the cube root. For g() = sin or g() = 2 we must limit the omain to obtain an inverse function. Eercise. The inverse image has the following properties: g (R) = R For any set A, g (A c ) = g (A) c For any collection of sets {A λ ; λ Λ}, g ( λ A λ ) = λ g (A). As a consequence the mapping A P {g() A} = P { g (A)} satisfies the aioms of a probability. The associate probability µ g() is calle the istribution of g().
Discrete Ranom Variables For a iscrete ranom variable with probabiliity mass function f, then the probability mass function f Y for Y = g() is easy to write. f Y (y) = f (). g (y) Eample 2. Let be a uniform ranom variable on {, 2,... n}, i. e., f () = /n for each in the state space. Then Y = + a is a uniform ranom variable on {a +, 2,... a + n} Eample 3. Let be a uniform ranom variable on { n, n +,..., n, n}. Then Y = has mass function { f Y (y) = 2n+ if = 0, 2 2n+ if 0. 2 Continuous Ranom Variable The easiest case for transformations of continuous ranom variables is the case of g one-to-one. We first consier the case of g increasing on the range of the ranom variable. In this case, g is also an increasing function. To compute the cumulative istribution of Y = g() in terms of the cumulative istribution of, note that F Y (y) = P {Y y} = P {g() y} = P { g (y)} = F (g (y)). Now use the chain rule to compute the ensity of Y For g ecreasing on the range of, an the ensity f Y (y) = F Y (y) = y F (g (y)) = f (g (y)) y g (y). F Y (y) = P {Y y} = P {g() y} = P { g (y)} = F (g (y)), f Y (y) = F Y (y) = y F (g (y)) = f (g (y)) y g (y). For g ecreasing, we also have g ecreasing an consequently the ensity of Y is inee positive, We can combine these two cases to obtain f Y (y) = f (g (y)) y g (y). Eample 4. Let U be a uniform ranom variable on [0, ] an let g(u) = u. Then g (v) = v, an V = U has ensity f V (v) = f U ( v) = on the interval [0, ] an 0 otherwise. 2
Eample 5. Let be a ranom variable that has a uniform ensity on [0, ]. Its ensity 0 if < 0, f () = if 0, 0 if >. Let g() = p, p 0. Then, the range of g is [0, ] an g (y) = y /p. If p > 0, then g is increasing an y g (y) = This ensity is unboune near zero whenever p >. If p < 0, then g is ecreasing. Its range is [, ), an y g (y) = 0 if y < 0, p y/p if 0 y, 0 if y >. { 0 if y <, p y/p if y, In this case, Y is a Pareto istribution with = an β = /p. We can obtain a Pareto istribution with arbitrary an β by taking ( ) /β g() =. If the transform g is not one-to-one then special care is necessary to fin the ensity of Y = g(). For eample if we take g() = 2, then g (y) = y. Thus, F y (y) = P {Y y} = P { 2 y} = P { y y} = F ( y) F ( y). f Y (y) = f ( y) y ( y) f ( y) y ( y) = 2 y (f ( y) + f ( y)) If the ensity f is symmetric about the origin, then f y (y) = y f ( y). Eample 6. A ranom variable Z is calle a stanar normal if its ensity is A calculus eercise yiels φ(z) = ep( z2 2 ). φ (z) = z ep( z2 2 ) = zφ(z), φ (z) = (z 2 ) ep( z2 2 ) = (z2 )φ(z). Thus, φ has a global maimum at z = 0, it is concave own if z < an concave up for z >. This show that the graph of φ has a bell shape. Y = Z 2 is calle a χ 2 (chi-square) ranom variable with one egree of freeom. Its ensity is f Y (y) = y ep( y 2 ). 3
3 The Probability Transform Let a continuous ranom variable whose istribution function F is strictly increasing on the possible values of. Then F has an inverse function. Let U = F (), then for u [0, ], P {U u} = P {F () u} = P {U F (u)} = F (F (u)) = u. In other wors, U is a uniform ranom variable on [0, ]. Most ranom number generators simulate inepenent copies of this ranom variable. Consequently, we can simulate inepenent ranom variables having istribution function F by simulating U, a uniform ranom variable on [0, ], an then taking = F (U). Eample 7. Let be uniform on the interval [a, b], then 0 if < a, a F () = b a if a b, if > b. Then u = a b a, (b a)u + a = = F (u). Eample 8. Let T be an eponential ranom variable. Thus, { 0 if t < 0, F T (t) = ep( t/β) if t 0. Then, u = ep( t/β), ep( t/β) = u, t = log( u). β Recall that if U is a uniform ranom variable on [0, ], then so is V = U. Thus if V is a uniform ranom variable on [0, ], then is a ranom variable with istribution function F T. Eample 9. Because β β t β+ T = β log V t = β t β A Pareto ranom variable has istribution function ( ) β =. Now, { 0 if <, F () = ( ) β if. ( ) β ( ) β u = u =, = ( u). /β 4
As before if V = U is a uniform ranom variable on [0, ], then = V /β is a Pareto ranom variable with istribution function F. 5