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jwimberley
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No.

For example, if $X$ follows a log normal distribution, where $\log(X) \sim N(\mu,\sigma)$, then $E[\log(x)] = \mu$ and is independent of $\sigma$. However, its mean is $E[X] = \exp \left(\mu + \frac{\sigma^2}{2} \right)$. Clearly, you cannot derive a $\sigma$ dependent number formfrom a $\sigma$ independent number.

No.

For example, if $X$ follows a log normal distribution, where $\log(X) \sim N(\mu,\sigma)$, then $E[\log(x)] = \mu$ and is independent of $\sigma$. However, its mean is $E[X] = \exp \left(\mu + \frac{\sigma^2}{2} \right)$. Clearly, you cannot derive a $\sigma$ dependent number form a $\sigma$ independent number.

No.

For example, if $X$ follows a log normal distribution, where $\log(X) \sim N(\mu,\sigma)$, then $E[\log(x)] = \mu$ and is independent of $\sigma$. However, its mean is $E[X] = \exp \left(\mu + \frac{\sigma^2}{2} \right)$. Clearly, you cannot derive a $\sigma$ dependent number from a $\sigma$ independent number.

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jwimberley
  • 4k
  • 2
  • 14
  • 20

No.

For example, if $X$ follows a log normal distribution, where $\log(X) \sim N(\mu,\sigma)$, then $E[\log(x)] = \mu$ and is independent of $\sigma$. However, its mean is $E[X] = \exp \left(\mu + \frac{\sigma^2}{2} \right)$. Clearly, you cannot derive a $\sigma$ dependent number form a $\sigma$ independent number.