# How to show probability satisfies probit model

This is an old past paper question that I am struggling to understand, so any help or hints would be appreciated...

Consider the choice between two options, such as two product brands. Let $U_0$ denote the utility of outcome $y = 0$ and $U_1$ the utility of $y = 1$. For $y = 0$ and $1$, suppose that $U_y = \alpha_y +\beta_yx+\epsilon_y$, using a scale such that εy has some standard distribution. A subject selects $y = 1$ if $U_1 > U_0$ for that subject. If $\epsilon_0$ and $\epsilon_1$ are independent $N(0,1)$ random variables, show that $P(Y = 1)$ satisfies the probit model.

So far, I have that $P(y=1)=P(U_1>U_0)=P(\alpha_1 +\beta_1x+\epsilon_1>\alpha_0 +\beta_0x+\epsilon_0)=P(\epsilon_0-\epsilon_1<\alpha_1-\alpha_0+x(\beta_1-\beta_0))$

but I am not sure where to go from here, or if what I am doing is even correct. Any help would be appreciated.

• Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. May 17 '16 at 15:08

$P(Y=1|X)=\Phi(X^T\beta),$
where $\Phi$ is the CDF of the normal. You're basically done since $\epsilon_0-\epsilon_1$ is going to be normally distributed with mean 0 and some variance like $2\sigma^2$. Then $P(\epsilon_1-\epsilon_2<X^T\beta)$, which after rescaling $\beta$ will give you $\Phi(X^T\beta)$.
• what do you mean by rescaling $\beta$ ? because the $\epsilon$s have variance 1, does that mean $\epsilon_0-\epsilon_1$ will have variance 2, so do I need to divide $\beta$ by 2? May 17 '16 at 18:55