# Proving That a Version of the Law of Total Probability Follows from Adam's Law

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.

• Quite fine your argument, there is no problem with the characteristic functions. You might want to have a look at en.wikipedia.org/wiki/Conditional_expectation for some generalizations of what you proved. – John B Nov 20 '18 at 21:59

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, $$\mathsf E(\mathrm I_A\mid X)$$, and you actually want to use the function of $$x$$, $$\mathsf E(\mathrm I_A\mid X{=}x)$$ inside the integral.

\begin{align}\mathsf P(A) &= \mathsf E(\mathrm I_A)\\&=\mathsf E(\mathsf E(\mathrm I_A\mid X))\\& = \int_{-\infty}^{\infty} \mathsf E(\mathrm I_A\mid 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(A\mid X{=}x)\,f_X(x)~\mathsf dx \end{align}

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