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Glen_b
Glen_b
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The $\mu$ and $\sigma^2$ parameters are the population mean and variance of the logs of the lognormal random variable with those parameters.

Your equations for them are correct - they're how the population mean and variance of the lognormal relate to the mean and variance of the log-variable.

Equating those expressions to the sample mean and variance would be a reasonable thing to do --- indeed, it's essentially method-of-moments$^\dagger$.

Those equations are rather straightforward to solve.

Divide the variance by the square of the mean, you get an equation in only $\sigma^2$ (one that's easily solved).

Then once you have solved that to get an estimate of $\sigma^2$, it's simple to substitute it back into the first equation to solve for your estimate of $\mu$.

If you want explicit formulas, see here or here$^\dagger$

$\dagger$ keeping in mind that the variance is a central moment but is readily obtained from the second raw moment and the mean, so equating sample and population raw moments should be equivalent to equating sample and population central moments (above the first) as long as you use the $n-$divisor versions of the central sample moments.

The $\mu$ and $\sigma^2$ parameters are the population mean and variance of the logs of the lognormal random variable with those parameters.

Your equations for them are correct - they're how the population mean and variance of the lognormal relate to the mean and variance of the log-variable.

Equating those expressions to the sample mean and variance would be a reasonable thing to do --- indeed, it's essentially method-of-moments.

Those equations are rather straightforward to solve.

Divide the variance by the square of the mean, you get an equation in only $\sigma^2$ (one that's easily solved).

Then once you have solved that to get an estimate of $\sigma^2$, it's simple to substitute it back into the first equation to solve for your estimate of $\mu$.

If you want explicit formulas, see here

The $\mu$ and $\sigma^2$ parameters are the population mean and variance of the logs of the lognormal random variable with those parameters.

Your equations for them are correct - they're how the population mean and variance of the lognormal relate to the mean and variance of the log-variable.

Equating those expressions to the sample mean and variance would be a reasonable thing to do --- indeed, it's essentially method-of-moments$^\dagger$.

Those equations are rather straightforward to solve.

Divide the variance by the square of the mean, you get an equation in only $\sigma^2$ (one that's easily solved).

Then once you have solved that to get an estimate of $\sigma^2$, it's simple to substitute it back into the first equation to solve for your estimate of $\mu$.

If you want explicit formulas, see here or here$^\dagger$

$\dagger$ keeping in mind that the variance is a central moment but is readily obtained from the second raw moment and the mean, so equating sample and population raw moments should be equivalent to equating sample and population central moments (above the first) as long as you use the $n-$divisor versions of the central sample moments.

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Glen_b
Glen_b
  • 290.5k
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The $\mu$ and $\sigma^2$ parameters are the population mean and variance of the logs of the lognormallognormal random variable with those parameters.

Your equations for them are correct - they're how the population mean and variance of the lognormal relate to the mean and variance of the log-variable.

Equating those expressions to the sample mean and variance would be a reasonable thing to do --- indeed, it's essentially method-of-momentsmethod-of-moments.

Those equations are rather straightforward to solve.

Divide the variance by the square of the mean, you get an equation in only $\sigma^2$ (one that's easily solved).

Then once you have solved that to get an estimate of $\sigma^2$, it's simple to substitute it back into the first equation to solve for your estimate of $\mu$.

If you want explicit formulas, see here

The $\mu$ and $\sigma^2$ parameters are the population mean and variance of the logs of the lognormal random variable with those parameters.

Your equations for them are correct - they're how the population mean and variance of the lognormal relate to the mean and variance of the log-variable.

Equating those expressions to the sample mean and variance would be a reasonable thing to do --- essentially method-of-moments.

Those equations are rather straightforward to solve.

Divide the variance by the square of the mean, you get an equation in only $\sigma^2$ (one that's easily solved).

Then once you have solved that to get an estimate of $\sigma^2$, it's simple to substitute it back into the first equation to solve for your estimate of $\mu$.

The $\mu$ and $\sigma^2$ parameters are the population mean and variance of the logs of the lognormal random variable with those parameters.

Your equations for them are correct - they're how the population mean and variance of the lognormal relate to the mean and variance of the log-variable.

Equating those expressions to the sample mean and variance would be a reasonable thing to do --- indeed, it's essentially method-of-moments.

Those equations are rather straightforward to solve.

Divide the variance by the square of the mean, you get an equation in only $\sigma^2$ (one that's easily solved).

Then once you have solved that to get an estimate of $\sigma^2$, it's simple to substitute it back into the first equation to solve for your estimate of $\mu$.

If you want explicit formulas, see here

Source Link
Glen_b
Glen_b
  • 290.5k
  • 37
  • 652
  • 1.1k

The $\mu$ and $\sigma^2$ parameters are the population mean and variance of the logs of the lognormal random variable with those parameters.

Your equations for them are correct - they're how the population mean and variance of the lognormal relate to the mean and variance of the log-variable.

Equating those expressions to the sample mean and variance would be a reasonable thing to do --- essentially method-of-moments.

Those equations are rather straightforward to solve.

Divide the variance by the square of the mean, you get an equation in only $\sigma^2$ (one that's easily solved).

Then once you have solved that to get an estimate of $\sigma^2$, it's simple to substitute it back into the first equation to solve for your estimate of $\mu$.