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I was speaking with someone today who told me that variance, in the sense of probability theory, is equivalent mathematically to energy in physics. Can anyone elaborate on this relationship?

rmh52
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2 Answers2

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In the Maxwell-Boltzmann kinetic theory of gases, a gas molecule has velocity modeled as a random vector $V$ whose components $V_x, V_y, V_z$ along three orthogonal axes are independent zero-mean Gaussian (or normal) random variable with variance $\sigma^2$. The kinetic energy of the particle is thus $\frac{1}{2}mV^2$ where $V$ is a random variable. Note that the expected value of the kinetic energy is thus $$E\left[\frac{1}{2}m|V|^2\right] = \frac{1}{2}mE[V_x^2+ V_y^2+ V_z^2] = \frac{3}{2}m\sigma^2.$$

Dilip Sarwate
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If you look at the formula of the variance, it is the same formula of calculating the power of a signal with its mean removed.

So, it is not energy from physics. It is power from signals, which don't necessarily have real world units.

In some cases (like telecommunications), this power from signal theory can be related with real phenomena and with real world power (watts) and real world energy (joules).

For more information I suggest reading about: