# Probit with variance not equal to 1

In probit models the latent variable is assumed to be of the form: $$y=\alpha+\beta x+\epsilon$$ with $$\epsilon \sim N(0,1)$$.

What if instead $$\epsilon \sim N(0,\sigma^2)$$?

Is there a way to estimate $$\sigma^2$$? Does it make a big difference to the model?

Assuming that $$\epsilon \sim \mathcal{N}(0,\sigma^2)$$ is the same with $$Y*:=y=\alpha+\beta x+\sigma\epsilon$$ where $$\epsilon \sim \mathcal{N}(0,1)$$
Then $$\mathbb{P}[Y=1|X] = \mathbb{P}[Y^*>0] = \mathbb{P}[\epsilon<\tfrac{1}{\sigma}(\alpha+\beta x)] = \Phi\left(\frac{\alpha+\beta x}{\sigma}\right) = \Phi\left(\tfrac{\alpha}{\sigma}+\tfrac{\beta}{\sigma}x\right) = \Phi\left(\tilde{\alpha}+\tilde{\beta}x\right)$$.
A model with $$\sigma^2$$ as a parameter to be estimated is not identified. We sent $$\sigma^2=1$$ simply to identify the model. In reality, we are estimating $$\frac{\beta_k}{\sigma}$$ for each $$k$$. See Breen, Karlson, & Holm (2018) for more details.