Suppose $Y=\beta_1 X + \beta_2 X^2$ for some real numbers $\beta_1$ and $\beta_2$ where $X$ is a random variable with real values greater than $0$ and $Y$ is greater than $0$.

What is the probability $P[Y>250 | X]$ if you consider $Y$ in a binary way as being either $>250$ or $<250$?

(This is a simplified version of equation $(7.3)$ in Introduction to Statistical Learning by Witten and Hastie on page $268$.)

I think we have that $P[Y>250 | X]=1-P[Y\leq 250 | X]$.

The text seems to claim that such a probability would be $\frac{e^{\beta_1 X + \beta_2 X^2}}{1+e^{\beta_1 X + \beta_2 X^2}}$ but I'm not following.

  • $\begingroup$ I don't get the question. $Y$ is $X$ measurable, hence $\mathbb P( \{Y> 250\} | X) := \mathbb E[ 1_{(250,+\infty)}(Y) | X] = 1_{(250,+\infty)}(Y)$. Note that conditional expectation/probability is a random variable and $\mathbb E[Z|\mathcal G] = Z$ if $Z$ is $\mathcal G$ measurable. $\endgroup$ Oct 7, 2020 at 10:55
  • $\begingroup$ The textbook claims that $P[Y>250|X]=\frac{e^{\beta_1 X + \beta_2 X^2}}{1+e^{\beta_1 X + \beta_2 X^2}}$ and that they treat $Y$ as binary in that of all the values of $Y$ that are possible, they separate it as those that are greater than or less than 250. I just don't see how they got their answer. $\endgroup$
    – E2R0NS
    Oct 7, 2020 at 11:18
  • $\begingroup$ In your notation what does $1_{(250, +\infty)}(Y)$ mean? $\endgroup$
    – E2R0NS
    Oct 7, 2020 at 11:19
  • $\begingroup$ $1_A$ stands for indicator function, that is $1_A(x)=1$ if $x \in A$ and $1_A(x)=0$ otherwise $\endgroup$ Oct 7, 2020 at 11:20
  • $\begingroup$ What resource are you using? $\endgroup$
    – E2R0NS
    Oct 7, 2020 at 11:22


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