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Carlo Beenakker
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Because of the central-limit-theorem, for large $N$ the absolute distance $d_N$ converges in distribution as $$P_N(d_N/\sqrt N)\to p(|X|),$$ where $X$ is a Gaussian random variable with mean zero and variance $2p=1-r$. Since $\mathbb{E}(|X|)=2\sqrt{p/\pi}$ we conclude that $$\mathbb{E}(d_N)\to \sqrt\frac{2(1-r)N}{\pi}.$$ So this "lazy" random walk differs from the simple random walk by a rescaling of the number of steps by a factor $1-r$.

Because of the central-limit-theorem, for large $N$ the absolute distance $d_N$ converges in distribution as $$P_N(d_N/\sqrt N)\to p(|X|),$$ where $X$ is a Gaussian random variable with mean zero and variance $2p=1-r$. Since $\mathbb{E}(|X|)=2\sqrt{p/\pi}$ we conclude that $$\mathbb{E}(d_N)\to \sqrt\frac{2(1-r)N}{\pi}.$$

Because of the central-limit-theorem, for large $N$ the absolute distance $d_N$ converges in distribution as $$P_N(d_N/\sqrt N)\to p(|X|),$$ where $X$ is a Gaussian random variable with mean zero and variance $2p=1-r$. Since $\mathbb{E}(|X|)=2\sqrt{p/\pi}$ we conclude that $$\mathbb{E}(d_N)\to \sqrt\frac{2(1-r)N}{\pi}.$$ So this "lazy" random walk differs from the simple random walk by a rescaling of the number of steps by a factor $1-r$.

Source Link
Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

Because of the central-limit-theorem, for large $N$ the absolute distance $d_N$ converges in distribution as $$P_N(d_N/\sqrt N)\to p(|X|),$$ where $X$ is a Gaussian random variable with mean zero and variance $2p=1-r$. Since $\mathbb{E}(|X|)=2\sqrt{p/\pi}$ we conclude that $$\mathbb{E}(d_N)\to \sqrt\frac{2(1-r)N}{\pi}.$$