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jld
jld
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I think there is no such $f$ in general unless there's some specifics to your problem that I'm missing.

We need $\sigma^2 f'(\mu)^2$ to not depend on $\sigma^2$. To try to formalize this, I'll interpret "not depend" to mean $$ \frac{\partial}{\partial \sigma^2} [\sigma^2 f'(\mu)^2] = 0. $$ In general the product rule gives us this derivative as $$ f'(\mu)^2 + 2\sigma^2 f'(\mu)f''(\mu)\frac{\partial \mu}{\partial \sigma^2}. $$

In a typical Gaussian distribution $\mu$ and $\sigma^2$ are free to vary separately and neither is viewed as a function of the other, soi.e. $\frac{\partial \mu}{\partial \sigma^2} = 0$. Furthermore you write $f$ and $f'$ as only functions of $\mu$, and in general I think it'd be very artificial to have them depend on $\sigma$ since then $f$ isn't a statistic, so I'll also assume that $f$ and $f'$ do not have $\sigma$ in them (that's why I didn't use the multivariate chain rule as I would have if we have $\frac{\text d}{\text d\sigma^2} f(\sigma^2, \mu(\sigma^2))$).

This means the derivative reduces tois just $$ f'(\mu)^2 = 0 \implies f'(\mu) = 0 \text{ everywhere}. $$ But functions with $f'(\mu) = 0$ are excluded from the delta method by hypothesis so there are no valid functions that satisfy our requirements here.

I think there is no such $f$ in general unless there's some specifics to your problem that I'm missing.

We need $\sigma^2 f'(\mu)^2$ to not depend on $\sigma^2$. To try to formalize this, I'll interpret "not depend" to mean $$ \frac{\partial}{\partial \sigma^2} [\sigma^2 f'(\mu)^2] = 0. $$ In general the product rule gives us this derivative as $$ f'(\mu)^2 + 2\sigma^2 f'(\mu)f''(\mu)\frac{\partial \mu}{\partial \sigma^2}. $$

In a typical Gaussian distribution $\mu$ and $\sigma^2$ are free to vary separately and neither is viewed as a function of the other, so $\frac{\partial \mu}{\partial \sigma^2} = 0$. Furthermore you write $f$ and $f'$ as only functions of $\mu$, and in general I think it'd be very artificial to have them depend on $\sigma$ since then $f$ isn't a statistic, so I'll also assume that $f$ and $f'$ do not have $\sigma$ in them (that's why I didn't use the multivariate chain rule as I would have if we have $\frac{\text d}{\text d\sigma^2} f(\sigma^2, \mu(\sigma^2))$).

This means the derivative reduces to just $$ f'(\mu)^2 = 0 \implies f'(\mu) = 0 \text{ everywhere}. $$ But functions with $f'(\mu) = 0$ are excluded from the delta method by hypothesis so there are no valid functions that satisfy our requirements here.

I think there is no such $f$ in general unless there's some specifics to your problem that I'm missing.

We need $\sigma^2 f'(\mu)^2$ to not depend on $\sigma^2$. To try to formalize this, I'll interpret "not depend" to mean $$ \frac{\partial}{\partial \sigma^2} [\sigma^2 f'(\mu)^2] = 0. $$

In a typical Gaussian distribution $\mu$ and $\sigma^2$ are free to vary separately and neither is viewed as a function of the other, i.e. $\frac{\partial \mu}{\partial \sigma^2} = 0$. Furthermore you write $f$ and $f'$ as only functions of $\mu$, and in general I think it'd be very artificial to have them depend on $\sigma$ since then $f$ isn't a statistic, so I'll also assume that $f$ and $f'$ do not have $\sigma$ in them.

This means the derivative is just $$ f'(\mu)^2 = 0 \implies f'(\mu) = 0 \text{ everywhere}. $$ But functions with $f'(\mu) = 0$ are excluded from the delta method by hypothesis so there are no valid functions that satisfy our requirements here.

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jld
jld
  • 20.8k
  • 2
  • 64
  • 69

I think there is no such $f$ in general unless there's some specifics to your problem that I'm missing.

We need $\sigma^2 f'(\mu)^2$ to not depend on $\sigma^2$. To try to formalize this, I'll interpret "not depend" to mean $$ \frac{\partial}{\partial \sigma^2} [\sigma^2 f'(\mu)^2] = 0. $$ In general the product rule gives us this derivative as $$ f'(\mu)^2 + 2\sigma^2 f'(\mu)f''(\mu)\frac{\partial \mu}{\partial \sigma^2}. $$

In a typical Gaussian distribution $\mu$ and $\sigma^2$ are free to vary separately and neither is viewed as a function of the other, so $\frac{\partial \mu}{\partial \sigma^2} = 0$. Furthermore you write $f$ and $f'$ as only functions of $\mu$, and in general I think it'd be very artificial to have them depend on $\sigma$ since then $f$ isn't a statistic, so I'll also assume that $f$ and $f'$ do not have $\sigma$ in them (that's why I didn't use the multivariate chain rule as I would have if we have $\frac{\text d}{\text d\sigma^2} f(\sigma^2, \mu(\sigma^2))$).

This means the derivative reduces to just $$ f'(\mu)^2 = 0 \implies f'(\mu) = 0 \text{ everywhere}. $$ But functions with $f'(\mu) = 0$ are excluded from the delta method by hypothesis so there are no valid functions that satisfy our requirements here.