Independent sums via characteristic functions

The Bernoulli transform is $1-p+p{e}^{it}$. Binomial weights $\left(\genfrac{}{}{0px}{}{n}{k}\right)p^k(1-p{)}^{n-k}$ are nonnegative and sum to one by the binomial theorem; they define a weighted Dirac probability, and the same finite expansion gives transform $(1-p+p{e}^{it}{)}^n$. Here every zeroth power means the empty product one, including parameter endpoints. The product rule for the given independent variables yields exactly this transform for their sum, so uniqueness gives its binomial law.

By [F22], $\mathrm{\mid }{e}^{itx}\mathrm{\mid }=1$ for real $t,x$. For $a\ge 0$, the weights $e^{-a}{a}^k\mathrm{/}k!$ sum to $e^{-a}{e}^a=1$ and define a probability. Truncating its exponential integrand to the integers $0,\dots ,N$ gives bounded functions of modulus at most one converging almost everywhere, so DCT identifies the transform with the absolutely convergent series $e^{-a}{\sum }_{k\ge 0}\frac{(a{e}^{it}{)}^k}{k!}=\exp (a(e^{it}-1))\mathrm{.}$ Multiplying the transforms at a=lambda and a=eta gives $\exp ((\lambda +\eta )(e^{it}-1))$, the transform of the constructed Poisson$(\lambda +\eta )$ law. Independence and uniqueness establish the claim, also when either parameter is zero.

For the normal calculation set $g(x)=e^{-x^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$, which has mass one. Its derivative is $-xg(x)$. On each half of $[-R,R]$, FTC gives ${\int }_{-R}^R\mathrm{\mid }x\mathrm{\mid }g(x)dx=2(1-e^{-R^2\mathrm{/}2})\mathrm{/}\sqrt{2\pi }$. The bridge and monotone convergence prove the first absolute moment finite. If a complex test $h=u+iv$ satisfies $\int \mathrm{\mid }h\mathrm{\mid }gdx<\mathrm{\infty }$, apply [F18] to the four nonnegative functions $u^{+},u^{-},v^{+},v^{-}$ and reassemble the finite integrals componentwise; hence $\int hd(gdx)=\int hgdx$. Applying this first to $h(x)=e^{itx}$ and then to $h(x)=ix{e}^{itx}$, whose absolute values are $1$ and $\mathrm{\mid }x\mathrm{\mid }$, the moments lemma gives ${\phi }^{\mathrm{\prime }}(t)=\int ix{e}^{itx}g(x)dx$. For fixed real $t$, [F22] writes $e^{itx}=\cos (tx)+i\sin (tx)$; [F14, F15, F23] therefore give $(d\mathrm{/}dx)e^{itx}=-t\sin (tx)+it\cos (tx)=it{e}^{itx}$, also for $t=0$. Compact integration by parts in both real components gives ${\int }_{-R}^{R}ix{e}^{itx}g(x)dx=-i[e^{itx}g(x){]}_{-R}^R-t{\int }_{-R}^R{e}^{itx}g(x)dx\mathrm{.}$ Both differentiated functions have continuous derivatives on the compact interval. The boundary is bounded by $2g(R)\to 0$, while $\mathrm{\mid }x\mathrm{\mid }g$ and g dominate the integrands; DCT yields ${\phi }^{\mathrm{\prime }}=-t\phi$. The real and imaginary derivatives of $e^{t^2\mathrm{/}2}\phi (t)$ are therefore zero by the product and chain rules, so both components are constant. Since $\phi (0)=1$, $\phi (t)=e^{-t^2\mathrm{/}2}$.

The normal definition and affine identity now give transform $e^{i{m}_{j}t-{\sigma }_j^2{t}^2\mathrm{/}2}$ for each input. Its product is $\exp (it{\sum }_j{m}_j-t^2{\sum }_j{\sigma }_j^2\mathrm{/}2)$, exactly the transform of $N({\sum }_j{m}_j,{\sum }_j{\sigma }_j^2)$; the nonnegative square root of the variance sum is the scale in that definition. Uniqueness proves the result. If all variances vanish, each input is constant and the result is the corresponding Dirac law; empty sums give ${\delta }_0$, and one-term sums return the original law. Bernoulli p=0 and p=1 similarly give deterministic zero and n. AC is inherited from Fourier uniqueness and from normal normalization and the compact integral bridge; no companion example is a supplier.