Characteristic Function

The characteristic function Similar to the moment generating function, we will define a function in terms of the exponential of a variable. Definition 0.1. Let X be a random variable. The characteristic function φX (t) is φX (t) := E(eitX ) For a continuous random variable and a discrete random variable, this is ∫ ∞ φX (t) = eitx fX (x)dx φX (t) = respectively. Note that the characteristic function for a continuous random variable is the complex conjugate to the Fourier transform of its density function. This fact implies that φX (t) determines fX (x) uniquely, and vice-versa. Indeed, the same holds for discrete random variables, but we will not show it. Example 0.2. Let Xi measure success on the ith trial of a BTP. Then E(eitX ) = qe0 + peit = q + peit . Example 0.3. Let X have the exponential density. Then the characteristic function is ∫ ∞ φX (t) = λeitx e−λx dx

0

∑

x

−∞

eitx P (X = x),

=

λ . λ − it

Example 0.4. Let X have the normal distribution with µ = 0 and σ = 1. Then ∫ ∞ 2 φX (t) = eitx e−x /2 dx.

−∞

We complete the square to obtain the exponent 1 (x − it)2 + 1 t2 , so 2 2 ∫ ∞ 2 1 1 2 φX (t) = e− 2 (x+it) − 2 t dx

−∞

2

= e−t =e

∫

/2

∞ −∞

e 2 y dy

1

2

−t2 /2

.

Here, I have been slightly cavalier about my change of variables, since we do not have complex analysis at our disposal, you will just have to trust me. Computing the characteristic function of a binomially distributed variable, say the number of successes in a BTP, would be more complicated. Fortunately, we can use the following fact to compute it instantly. Proposition 0.5. Let X and Y be independent random variables. Then φX+Y (t) = φX (t)φY (t). Proof. It follows by simply writing out the expression for the characteristic function: φX+Y (t) = E(eit(X+Y ) ) = E(eitX eitY ) = E(eitX )E(eitY ) = φX (t)φY (t).

In the penultimate line, we have used the fact that any functions g(X) and h(Y ) are independent since X and Y are. Corollary 0.6. For a...