Proving That a Version of the Law of Total Probability Follows from Adam's Law
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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.
probability conditional-expectation conditional-probability
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up vote
0
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favorite
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.
probability conditional-expectation conditional-probability
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
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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.
probability conditional-expectation conditional-probability
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.
probability conditional-expectation conditional-probability
probability conditional-expectation conditional-probability
asked yesterday
JStorm
255
255
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
yesterday
add a comment |
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
yesterday
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
yesterday
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
yesterday
add a comment |
1 Answer
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1
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accepted
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: $mathsf P(A)=mathsf E(mathsf P(Amid X))$ .
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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: $mathsf P(A)=mathsf E(mathsf P(Amid X))$ .
add a comment |
up vote
1
down vote
accepted
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: $mathsf P(A)=mathsf E(mathsf P(Amid X))$ .
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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: $mathsf P(A)=mathsf E(mathsf P(Amid X))$ .
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: $mathsf P(A)=mathsf E(mathsf P(Amid X))$ .
answered yesterday
Graham Kemp
84k43378
84k43378
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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
yesterday