Normally distributed X implies logistic regression for Y

**noob mathematician** · March 23rd 2011, 03:36 AM

Suppose the distribution of X for subjects having Y=1 is normal $N(\mu_1,\sigma)$ , and suppose the distribution of X for subjects having Y=0 is normal $N(\mu_0,\sigma)$ . Then why is $P(Y=1|x)$ a logistic regression curve?

And why $\beta=(\mu_1-\mu_0/\sigma^2)$ ?

**theodds** · March 23rd 2011, 09:52 AM

Note that Y is Bernoulli with probability $\pi$ (say) and we are given that $X|Y \sim N(\mu_Y, \sigma^2)$ . Just calculate

$f_{Y|X} (1 | x) = \frac{f_{X|Y} (x|1) f_Y (1)}{f_{X|Y} (x|0) f_Y (0) + f_{X|Y} (x|1) f_Y (1)}$

and see what you get. I haven't done it all the way through, but it looks to me like you will get a function whose inverse is the logit.

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