4 deleted 16 characters in body edited Jul 4 '18 at 22:47 Ben 33.2k22 gold badges3838 silver badges146146 bronze badges Unless theThe standard deviation is zero, it is strictly greater thanat least as large as the mean absolute deviation: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values; from Jensen's inequality it follows that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(|X-\mu|)=0$$. It follows that $$\mathbb{S}(X) \geqslant \mathbb{V}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{V}(|X-\mu|)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation. Unless the standard deviation is zero, it is strictly greater than mean absolute deviation: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values; from Jensen's inequality it follows that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(|X-\mu|)=0$$. It follows that $$\mathbb{S}(X) \geqslant \mathbb{V}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{V}(|X-\mu|)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation. The standard deviation is at least as large as the mean absolute deviation: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values; from Jensen's inequality it follows that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(|X-\mu|)=0$$. It follows that $$\mathbb{S}(X) \geqslant \mathbb{V}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{V}(|X-\mu|)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation. 3 Corrected condition for strict inequality edited Jul 4 '18 at 13:49 Ben 33.2k22 gold badges3838 silver badges146146 bronze badges Unless the standard deviation is zero, it is strictly greater than mean absolute deviation: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values; from Jensen's inequality it follows that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(X)=0$$$$\mathbb{V}(|X-\mu|)=0$$. It therefore follows that $$\mathbb{S}(X) \geqslant \mathbb{E}(|X-\mu|)$$$$\mathbb{S}(X) \geqslant \mathbb{V}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{S}(X)=0$$$$\mathbb{V}(|X-\mu|)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation. Unless the standard deviation is zero, it is strictly greater than mean absolute deviation: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values; from Jensen's inequality it follows that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(X)=0$$. It therefore follows that $$\mathbb{S}(X) \geqslant \mathbb{E}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{S}(X)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation. Unless the standard deviation is zero, it is strictly greater than mean absolute deviation: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values; from Jensen's inequality it follows that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(|X-\mu|)=0$$. It follows that $$\mathbb{S}(X) \geqslant \mathbb{V}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{V}(|X-\mu|)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation. 2 edited body edited Jul 2 '18 at 7:21 Ben 33.2k22 gold badges3838 silver badges146146 bronze badges Unless the standard deviation is zero, it is strictly greater than MABmean absolute deviation: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values, so it followsvalues; from Jensen's inequality it follows that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(X)=0$$. It therefore follows that $$\mathbb{S}(X) \geqslant \mathbb{E}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{S}(X)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation (MAB). Unless the standard deviation is zero, it is strictly greater than MAB: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values, so it follows from Jensen's inequality that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(X)=0$$. It therefore follows that $$\mathbb{S}(X) \geqslant \mathbb{E}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{S}(X)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation (MAB). Unless the standard deviation is zero, it is strictly greater than mean absolute deviation: The function $$\varphi(z) = z^2$$ is strictly convex over non-negative argument values; from Jensen's inequality it follows that: $$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$ with the inequality being strict unless $$\mathbb{V}(X)=0$$. It therefore follows that $$\mathbb{S}(X) \geqslant \mathbb{E}(|X-\mu|)$$, with the inequality being strict unless $$\mathbb{S}(X)=0$$. So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation. 1 answered Jul 2 '18 at 7:10 Ben 33.2k22 gold badges3838 silver badges146146 bronze badges