Jtdylktuy

I have a homework question that asks:

Show that the following version of LOTP follows from Adam’s law: for any event A and continuous random variable X with PDF $f_X$:

$$ P(A) = int_{- infty}^{infty} P(A|X=x)f_X(x) dx $$

[Edited to add: Adam's Law is also the Law of Total Expectation and the Law of Iterated Expectation, and my text gives it as: $E(Y) = E(E(Y|X))$]

Here is the proof I have written:

$P(A) = E(I_A)$ and $P(A|X = x) = E(I_A|X = x)$ by the fundamental bridge.

Let $E(I_A|X = x) = g(x)$, a function of x, then:

$E(g(X)) = int_{- infty}^{infty} g(x)f_X(x) dx $ (This is a formula I found in my text)

$= E(E(I_A|X)) = E(I_A)$ (by Adam's)

$= P(A)$ (by the bridge)

Therefore,

$P(A) = int_{- infty}^{infty} E(I_A|X)f_X(x) dx = int_{- infty}^{infty} P(A|X)f_X(x) dx $

My main concern is that in this proof, I have dropped the expected value of the indicator variable into the integral, but I believe indicator variables are always discrete. However I'm not sure how to cope with this, because I am asked to connect the given (continuous) formula to Adam's, which requires expectation, and connecting probability to expectation requires indicator variables.

Therefore,

$mathsf P(A) = int_{- infty}^{infty} mathsf E(I_A|X)f_X(x) dx = int_{- infty}^{infty} mathsf P(A|X)f_X(x) dx $

My main concern is that in this proof, I have dropped the expected value of the indicator variable into the integral, but I believe indicator variables are always discrete.

It is not a concern.

You are not using the indicator random variable, but its conditional exectation, $mathsf E(mathrm I_Amid X)$, and you actually want to use the function of $x$, $mathsf E(mathrm I_Amid X{=}x)$ inside the integral.

$$begin{align}mathsf P(A) &= mathsf E(mathrm I_A)\&=mathsf E(mathsf E(mathrm I_Amid X))\& = int_{-infty}^{infty} mathsf E(mathrm I_Amid X{=}x),f_X(x)~mathsf dx &~:~& mathsf E(g(X))=int_Bbb R g(x)~f_X(x)~mathsf d x \ & = int_{-infty}^{infty} mathsf P(Amid X{=}x),f_X(x)~mathsf dx end{align}$$

In short the Law of Total Probability Is: $mathsf P(A)=mathsf E(mathsf P(Amid X))$ .