Solution: Computing CFs of Standard Distributions by Integration

Exercise: Computing CFs of Standard Distributions by Integration

Part 1 — Normal

\varphi_X(t) = \frac{1}{\sigma\sqrt{2\pi}}\int_{\mathbb{R}} e^{itx}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx.

Complete the square in $x$ :

itx - \frac{(x-\mu)^2}{2\sigma^2} = -\frac{1}{2\sigma^2}\left(x - (\mu + it\sigma^2)\right)^2 + i\mu t - \tfrac{1}{2}\sigma^2 t^2.

Pulling the constant outside and completing the contour shift (which is valid because the Gaussian decays faster than any exponential), the remaining integral equals $\sigma\sqrt{2\pi}$ . Hence:

\boxed{\varphi_X(t) = \exp\!\left(i\mu t - \tfrac{1}{2}\sigma^2 t^2\right).}

Part 2 — Exponential

\varphi_X(t) = \int_0^\infty \lambda e^{-(\lambda - it)x}\,dx = \frac{\lambda}{\lambda - it} \quad\text{for all } t \in \mathbb{R}.

The integral converges because $\operatorname{Re}(\lambda - it) = \lambda > 0$ regardless of $t$ . So $\boxed{\varphi_X(t) = \lambda/(\lambda - it)}$ .

Part 3 — Poisson

\varphi_X(t) = \sum_{k=0}^\infty e^{itk}\cdot \frac{e^{-\lambda}\lambda^k}{k!} = e^{-\lambda}\sum_{k=0}^\infty \frac{(\lambda e^{it})^k}{k!} = e^{-\lambda}\cdot e^{\lambda e^{it}} = \boxed{\exp(\lambda(e^{it} - 1)).}

Part 4 — Poisson convolution

For $X_i \sim \text{Poisson}(\lambda_i)$ independent:

\varphi_{X_1 + X_2}(t) = \varphi_{X_1}(t)\varphi_{X_2}(t) = e^{\lambda_1(e^{it}-1)}\cdot e^{\lambda_2(e^{it}-1)} = e^{(\lambda_1 + \lambda_2)(e^{it}-1)}.

This is the CF of a $\text{Poisson}(\lambda_1 + \lambda_2)$ . By CF uniqueness, $X_1 + X_2 \sim \text{Poisson}(\lambda_1 + \lambda_2)$ — the well-known Poisson-convolution rule, recovered as one line of CF algebra.

Takeaways

CFs convert density integrals into clean algebraic formulae. Memorise the Gaussian CF $e^{i\mu t - \sigma^2 t^2/2}$ — it appears everywhere.
CF multiplicativity encodes the convolution rule for sums of independents. Many "closed under addition" properties (normal, Poisson, gamma with same rate) are one-line CF arguments.
Complex-analysis contour shifts are the routine trick for Gaussian-type CFs. Beyond sketchable here, but good to be aware of when deriving CFs of more general distributions.