Consider the gaussian distribution
ρ(x) = A e^{-λ(x-a)^2} ,
where A, a, and are positive real constants. (The necessary integrals are inside the back cover.) .
(a) Use Equation 1.16 to determine A.
(b) Find \left\langle x\right\rangle , \left\langle x^2\right\rangle , and σ.
(c) Sketch the graph of ρ(x).
\int_{0}^{h}{\frac{1}{2\sqrt{hx} } } dx=\frac{1}{2\sqrt{h} }\Bigl(2x^{1/2}\Bigr)\mid ^h_0=1 (1.16).
(a)
1=\int_{-\infty }^{\infty }{Ae^{-\lambda (x-a)^2}} dx . Let u ≡ x-a , du = dx , u : -\infty → \infty .
1=A\int_{-\infty }^{\infty } {e^{-λu^2}}du = A \sqrt{\frac{π}{λ} } ⇒ A = \sqrt{\frac{λ}{π} } .
(b)
\left\langle x\right\rangle = \int_{-\infty }^{\infty } {x^{-λ(x-a)^2}}dx = A \int_{-\infty }^{\infty } {(u+a)e^{-λu^2}}du .
= A \biggl[ \int_{-\infty }^{\infty }{ue^{-λu^2}}du + a \int_{-\infty }^{\infty }{e^{-λu^2}}du\biggr] = A \biggl(0+a \sqrt{\frac{π}{λ} } \biggr) = a .
\left\langle x^2\right\rangle = A\int_{-\infty }^{\infty }{x^2 e^{-λ(x-a)^2}}dx .
= A \left\{\int_{-\infty }^{\infty }{u^2 e^{-λu^2}} du + 2a\int_{-\infty }^{\infty }{u e^{-λu^2}} du + a^2 \int_{-\infty }^{\infty }{ e^{-λu^2}} du\right\} .
= A \biggl[\frac{1}{2λ} \sqrt{\frac{π}{λ} } + 0 + a^2 \sqrt{\frac{π}{λ} } \biggr] = a^2 + \frac{1}{2λ} .
σ^2 = \left\langle x^2\right\rangle – \left\langle x\right\rangle^2 = a^2 + \frac{1}{2λ} – a^2 = \frac{1}{2λ} .
σ = \frac{1}{ \sqrt{2λ}} .
(c)
