Consider the random geometric graph G=G(n,rn,f)\documentclass[12pt]{minimal}
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\begin{document}$$G = G(n,r_n,f)$$\end{document} consisting of n nodes independently distributed in S=-12,122\documentclass[12pt]{minimal}
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\begin{document}$$S = \left[ -\frac{1}{2},\frac{1}{2}\right] ^2$$\end{document} each according to a density f(·)\documentclass[12pt]{minimal}
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\begin{document}$$f({\cdot })$$\end{document}. Two nodes u and v are joined by an edge if the Euclidean distance between them is less than rn.\documentclass[12pt]{minimal}
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\begin{document}$$r_n.$$\end{document} Let CG\documentclass[12pt]{minimal}
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\begin{document}$$C_G$$\end{document} denote the component of G containing the largest number of nodes and denote diam(CG)\documentclass[12pt]{minimal}
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\begin{document}$$\text {diam}(C_G)$$\end{document} to be its diameter. Let s and t be the nodes of G closest to -12,12\documentclass[12pt]{minimal}
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\begin{document}$$\left( -\frac{1}{2},\frac{1}{2}\right) $$\end{document} and 12,12,\documentclass[12pt]{minimal}
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\begin{document}$$\left( \frac{1}{2},\frac{1}{2}\right) ,$$\end{document} respectively and let dG(s,t)\documentclass[12pt]{minimal}
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\begin{document}$$d_G(s,t)$$\end{document} denote their graph distance. We prove that the normalized diameter rn2diam(CG)\documentclass[12pt]{minimal}
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\begin{document}$$\frac{r_n}{\sqrt{2}} \text {diam}(C_G)$$\end{document} and the stretch rndG(s,t)\documentclass[12pt]{minimal}
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\begin{document}$$r_nd_G(s,t)$$\end{document} both converge to one in probability if nrn2→∞\documentclass[12pt]{minimal}
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\begin{document}$$nr_n^2 \rightarrow \infty $$\end{document} as n→∞\documentclass[12pt]{minimal}
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\begin{document}$$n \rightarrow \infty $$\end{document}.
