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I'm confused on how to apply the law of total probability when the summation is $\sum_{0}^{1}$ i.e. from $0$ to $1$, including all possible decimal numbers in between, rather than a summation of $\sum_{a}^{b}$ where $a$ and $b$ are integers. Is the following the correct approach (using an integral from $0$ to $1$)?

If I apply the law of total probability to $P(Y=1)$, does it equal:

$ \int_{0}^{1} P(Y=1|X=x)P(X=x)dx$

where $Y \in \{0,1\}$ and $x \in [0,1]$

TopoSet32
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ahkl789
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    What is the actual question? How can we tell if the approach is right if you don't tell us what you are trying to do? "Sum of an equation where the limits are 0 to 1" doesn't make any sense. – saulspatz Aug 22 '19 at 13:39

1 Answers1

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No.

If e.g. $X$ has continuous distribution then your integrand is $0$ for every $x$ hence also the integral takes value $0$.

Go for:

$$\begin{aligned}P\left(Y=1\right) & =\mathbb{E}\left[\mathbf{1}_{Y=1}\right]\\ & =\mathbb{E}\left[\mathbb{E}\left[\mathbf{1}_{Y=1}\mid X\right]\right]\\ & =\int\mathbb{E}\left[\mathbf{1}_{Y=1}\mid X=x\right]F_{X}\left(x\right)\\ & =\int P\left[Y=1\mid X=x\right]F_{X}\left(x\right) \end{aligned} $$

drhab
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  • If X does not have a continuous distribution, so P(X=x)>0, but still takes the values in the interval [0,1], could I use a summation in place of the integral in my question? – ahkl789 Aug 22 '19 at 15:20
  • If $X$ does not have a continuous distribution then still we get $\int_0^1P(Y=1\mid X=x)P(X=x)dx=0$ (what you mention in your question) and is wrong. If there is a countable set $S$ with $P(X\in S)=1$ then the integral in my answer will be equal to $\sum_{s\in S}P(Y=1\mid X=s)P(X=s)$. – drhab Aug 22 '19 at 18:31