Understanding formulas for $E(X∣X < c)$
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I have $E(X∣X < c)$ where $X$ is a continuous random variable and $c$ is a given positive real value.
According to this question this is equivalent to:
- $E[X|X<c]=int_{-inf}^{c}1-frac{F(x)}{F(c)}dx$
On this other question we have both:
- $mathrm E(Xmid X < c)=frac{mathrm E(Xcdotmathbf 1_{X < c})}{mathrm P(Xgt c)}$
- $E(X|X<c) = frac{int_x x f(x) I(x<c) dx}{int_x f(x) I(x<c) dx}$
Yet, starting from the definition of conditional expectation, I couldn't get to show any of the three equations.
I am particularly interested in the first one.
Can you help me understand how to get it? (And or the others)
Thanks.
probability conditional-expectation
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add a comment |
$begingroup$
I have $E(X∣X < c)$ where $X$ is a continuous random variable and $c$ is a given positive real value.
According to this question this is equivalent to:
- $E[X|X<c]=int_{-inf}^{c}1-frac{F(x)}{F(c)}dx$
On this other question we have both:
- $mathrm E(Xmid X < c)=frac{mathrm E(Xcdotmathbf 1_{X < c})}{mathrm P(Xgt c)}$
- $E(X|X<c) = frac{int_x x f(x) I(x<c) dx}{int_x f(x) I(x<c) dx}$
Yet, starting from the definition of conditional expectation, I couldn't get to show any of the three equations.
I am particularly interested in the first one.
Can you help me understand how to get it? (And or the others)
Thanks.
probability conditional-expectation
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$begingroup$
The first one looks almost right when you integrate by parts. The third one is a roundabout way of integrating from$-infty$ to $c$ for both the the numerator and the denominator. The second seems to have an error for the denominator.
$endgroup$
– herb steinberg
Dec 2 '18 at 23:27
$begingroup$
Ps. Ensure that you use the definition for expectation conditioned over an event, rather than a random variable.
$endgroup$
– Graham Kemp
Dec 3 '18 at 0:05
add a comment |
$begingroup$
I have $E(X∣X < c)$ where $X$ is a continuous random variable and $c$ is a given positive real value.
According to this question this is equivalent to:
- $E[X|X<c]=int_{-inf}^{c}1-frac{F(x)}{F(c)}dx$
On this other question we have both:
- $mathrm E(Xmid X < c)=frac{mathrm E(Xcdotmathbf 1_{X < c})}{mathrm P(Xgt c)}$
- $E(X|X<c) = frac{int_x x f(x) I(x<c) dx}{int_x f(x) I(x<c) dx}$
Yet, starting from the definition of conditional expectation, I couldn't get to show any of the three equations.
I am particularly interested in the first one.
Can you help me understand how to get it? (And or the others)
Thanks.
probability conditional-expectation
$endgroup$
I have $E(X∣X < c)$ where $X$ is a continuous random variable and $c$ is a given positive real value.
According to this question this is equivalent to:
- $E[X|X<c]=int_{-inf}^{c}1-frac{F(x)}{F(c)}dx$
On this other question we have both:
- $mathrm E(Xmid X < c)=frac{mathrm E(Xcdotmathbf 1_{X < c})}{mathrm P(Xgt c)}$
- $E(X|X<c) = frac{int_x x f(x) I(x<c) dx}{int_x f(x) I(x<c) dx}$
Yet, starting from the definition of conditional expectation, I couldn't get to show any of the three equations.
I am particularly interested in the first one.
Can you help me understand how to get it? (And or the others)
Thanks.
probability conditional-expectation
probability conditional-expectation
asked Dec 2 '18 at 21:51
jpuertajpuerta
11
11
$begingroup$
The first one looks almost right when you integrate by parts. The third one is a roundabout way of integrating from$-infty$ to $c$ for both the the numerator and the denominator. The second seems to have an error for the denominator.
$endgroup$
– herb steinberg
Dec 2 '18 at 23:27
$begingroup$
Ps. Ensure that you use the definition for expectation conditioned over an event, rather than a random variable.
$endgroup$
– Graham Kemp
Dec 3 '18 at 0:05
add a comment |
$begingroup$
The first one looks almost right when you integrate by parts. The third one is a roundabout way of integrating from$-infty$ to $c$ for both the the numerator and the denominator. The second seems to have an error for the denominator.
$endgroup$
– herb steinberg
Dec 2 '18 at 23:27
$begingroup$
Ps. Ensure that you use the definition for expectation conditioned over an event, rather than a random variable.
$endgroup$
– Graham Kemp
Dec 3 '18 at 0:05
$begingroup$
The first one looks almost right when you integrate by parts. The third one is a roundabout way of integrating from$-infty$ to $c$ for both the the numerator and the denominator. The second seems to have an error for the denominator.
$endgroup$
– herb steinberg
Dec 2 '18 at 23:27
$begingroup$
The first one looks almost right when you integrate by parts. The third one is a roundabout way of integrating from$-infty$ to $c$ for both the the numerator and the denominator. The second seems to have an error for the denominator.
$endgroup$
– herb steinberg
Dec 2 '18 at 23:27
$begingroup$
Ps. Ensure that you use the definition for expectation conditioned over an event, rather than a random variable.
$endgroup$
– Graham Kemp
Dec 3 '18 at 0:05
$begingroup$
Ps. Ensure that you use the definition for expectation conditioned over an event, rather than a random variable.
$endgroup$
– Graham Kemp
Dec 3 '18 at 0:05
add a comment |
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$begingroup$
The first one looks almost right when you integrate by parts. The third one is a roundabout way of integrating from$-infty$ to $c$ for both the the numerator and the denominator. The second seems to have an error for the denominator.
$endgroup$
– herb steinberg
Dec 2 '18 at 23:27
$begingroup$
Ps. Ensure that you use the definition for expectation conditioned over an event, rather than a random variable.
$endgroup$
– Graham Kemp
Dec 3 '18 at 0:05