# How to define an inverse gamma distribution with a fixed mode but a changeable variance for a bayesian prior?

I'm trying to define a prior distribution for a research project using bayesian estimation that's from a non-normally distributed posterior. Since it's not normally distributed I've been recommended to use an inverse gamma distribution. I have the mean, mode, and standard deviation of this distribution and have been recommended to define the prior using the same mode as the posterior but to increase the variance by 4x (so it's weakly informative). Here's a screenshot of the posterior distribution I'm using to define the prior:

It has mean = 0.286, Std = 0.157, and a mode = 0.153. Using the following equation from Appendix A of an Mplus guide (the program I'm using)

$$\alpha = 2 + \frac{\mu^2}{\sigma^2}$$

$$\beta = \mu + \frac{\mu^3}{\sigma^2}$$

I've been able to derive an $$IG(\alpha, \beta)$$ distribution of $$IG(4.45, 0.87)$$. Now I want to expand on that by increasing the variance by 4 times while keeping the mode the same. This has proved difficult due to my lack of experience in this field. The following are equations I've been given that might help calculate the mean, variance, and mode.

$$\mu = \frac{\beta}{(\alpha-1)}$$

$$\sigma^2 = \frac{\beta^2}{(\alpha-1)^2(\alpha-2)}$$

$$\text{Mode} = \frac{\beta}{(a+1)}$$

With these I haven't really been able to replicate the mean, variance, and mode based on the IG I defined earlier. Since IG is a cumulative distribution, I could use random $$\alpha$$ and $$\beta$$ values that increase variance until I land on a distribution with a wide enough variance, but this would be atheoretical.

Any help or advice on this is much appreciated since I'm a novice when it comes to bayesian priors and inverse gamma distributions.

If I understand you correctly you know the mode and variance your targeting (Mode and $$\sigma^2$$ are parameters). Using the formulas $$\beta=Mode (\alpha+1)$$. Plugging this in to the formula for variance.
$$\sigma^2 = \frac{Mode^2 (\alpha+1)^2}{(\alpha-1)^2(\alpha-2)}$$.
This gives a cubic equation in $$\alpha$$ you can solve numerically. Then you can solve $$\beta=Mode (\alpha+1)$$.