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Let $\phi : \mathbb{R}^n \rightarrow \mathbb{R}$ be the Gaussian radial basis function: $$\phi(x) = \exp(-|x|^2)$$

Let $$f_i(x) = \phi{\left(\frac{x - \mu_i}{\sigma_i}\right)}$$

I'm computing the inner product between $f_i$ and $f_j$. My calculation is as follows: \begin{align*} \langle f_i, f_j \rangle &= \int_{\mathbb{R}^n} \langle f_i(x), f_j(x) \rangle \omega \\ &= \int_{\mathbb{R}^n} \phi\left(\frac{x - \mu_i}{\sigma_i}\right) \phi\left(\frac{x - b_j}{\sigma_j}\right) \omega \\ &= \int_{\mathbb{R}^n} \phi\left(\frac{x - \mu_i}{\sigma_i}\right) \phi\left(\frac{x - b_j}{\sigma_j}\right) \omega \\ &= \int_{\mathbb{R}^n} \exp\left(-\left|\frac{x-\mu_i}{\sigma_i}\right|^2\right) \exp\left(-\left|\frac{x-\mu_j}{\sigma_j}\right|^2\right) \omega \\ &= \int_{\mathbb{R}^n} \exp\left(-\left(x-\frac{\mu_i\sigma_j^2+\mu_j\sigma_i^2}{\sigma_i^2+\sigma_j^2}\right)^2\left(\frac{1}{\sigma_i^2}+\frac{1}{\sigma_j^2}\right) - \frac{(b_i-b_j)^2}{\sigma_i^2+\sigma_j^2}\right) \omega \\ &= \exp\left(-\frac{(b_i-b_j)^2}{\sigma_i^2+\sigma_j^2}\right) \int_{\mathbb{R}^n} \exp\left(-x^2\left(\sigma_i^{-2}+\sigma_j^{-2}\right)\right) \omega \\ &= \exp\left(-\frac{(b_i-b_j)^2}{\sigma_i^2+\sigma_j^2}\right) \sqrt{\det 2\pi(\sigma_i^{-2}+\sigma_j^{-2})^{-1} I} \\ &= \exp\left(-\frac{(b_i-b_j)^2}{\sigma_i^2+\sigma_j^2}\right) (2\pi (\sigma_i^{-2}+\sigma_j^{-2})^{-1})^{n/2} \end{align*}

Is the final answer correct? Does anyone know of a source/paper that confirms this answer?

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