4 deleted 16 characters in body
source | link

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
source | link

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
source | link

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
source | link