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I'm having a bit of trouble understanding exactly why there is a "1" in the general simple Binary Choice Model where $Y_i$ can take a value of either $0$ or $1$. We also assume that the conditional distribution of $\epsilon$ given $X$ is a standard normal distribution.

For example, in my notes, our econometrics professor wrote the following:

$$E[Y_i|X_i=t]=P[B_0+B_1X_i\geq\epsilon_i|X_i=t]$$ and the next step was: $$E[Y_i|X_i=t]=E[1(B_0+B_1X_i\geq\epsilon_i|X_i=t]$$

My question is about this $1$ that keeps popping up. What is the significance of this? Also How can we go from the line above to the line below?

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    $\begingroup$ Although your expressions are syntactically incorrect (probably just some typographical errors in the markup), it's likely the "$1$" refers to the indicator function. $\endgroup$ – whuber Apr 14 '15 at 23:11
  • $\begingroup$ I see. But can you please explain how to go from one line to the other? Thank you. $\endgroup$ – nicefella Apr 15 '15 at 20:30
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    $\begingroup$ By definition, the expectation is a sum of values times probabilities. When the values are only $1$ and $0$, multiplication by $0$ introduces nothing while multiplication by $1$ leaves just the probabilities. $\endgroup$ – whuber Apr 15 '15 at 20:33
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As @whuber commented, the most plausible guess is that your $1$ refers to the indicator function.

You go from one line to the other by noting that the indicator function can only take two values: $1$ if the event it indicates is true and $0$ when it is false, so (again, just following @whuber's comment)

$E[I_{\{B_0+B_1X_i \geq\epsilon_i\}}|X_i=t] = $, by expectation definition for a discrete variable,

$\mathbb{P}(I_{\{B_0+B_1X_i \geq\epsilon_i\}}=1|X_i=t).1 + \mathbb{P}(I_{\{B_0+B_1X_i \geq\epsilon_i\}}=0|X_i=t).0 = $

$\mathbb{P}(I_{\{B_0+B_1X_i \geq\epsilon_i\}}=1|X_i=t)$

And since the events $\{I_{\{B_0+B_1X_i \geq\epsilon_i\}}=1\}$ and $\{B_0+B_1X_i \geq\epsilon_i\}$ are the same event (the indicator equals one if and only if the event on the right occurs), you have that:

$\mathbb{P}(I_{\{B_0+B_1X_i \geq\epsilon_i\}}=1|X_i=t) =$ $\mathbb{P}(B_0+B_1X_i \geq\epsilon_i|X_i=t)$

And the equality you were looking for.

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