What does the integral of a function times a function of a random variable represent, conceptually?Expected...
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What does the integral of a function times a function of a random variable represent, conceptually?
Expected value of a Gaussian random variable transformed with a logistic functionWhat is the expected partial value function really called?indicator variable - dirac delta or step functionExpected value of bounded function?Why does MLE not include the integral for joint probability of a contious random variableHow to calculate the expected value of a standard normal distribution?Plot the density function of a normal random variable knowing only the characteristic function in RExpectation of a function of a random variable from CDFDerivation of variance of normal distribution with gamma functionMoment Generating Function for Lognormal Random Variable
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$begingroup$
I am trying to understand conceptually what does the following give me or tell me:
$$int f(x) cdot g(x) , dx$$
where $f(x)$ is any continuous function of $x$ and $g(x)$ is the probability density function for a random variable, for example a normal distribution's PDF is:
$$ g(x) = frac1 {2 pi sigma^2} expleft(frac{- (x - mu)^2 }{ 2 sigma ^2}right) $$
I understand the integral of a PDF gives me the CDF. So:
$$int_{-infty}^0 g(x) , dx$$
Gives me the probability of $x$ being less than $0$. However, what happens when you multiply $g(x)$ by another function $f(x)$ and take the integral? I heard it gives you the expected payoff assuming $f(x)$ is a function of payoffs and you take an integral from -infinity to +infinity. This, if true, I conceptually understand. sum of payoffs times the probabilities is the expected value of whatever game you are playing.
I start getting confused when the boundaries of the integral are not $pm infty$. I'm not sure in that case what integral conceptually means. For example:
$$int_{-infty}^0 f(x) g(x) , dx$$
What does that tell me?
probability distributions normal-distribution random-variable expected-value
edited 5 hours ago
Siong Thye Goh
3,0842621
asked 5 hours ago
vt_ogvt_og
141
New contributor
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to understand conceptually what does the following give me or tell me:
$$int f(x) cdot g(x) , dx$$
where $f(x)$ is any continuous function of $x$ and $g(x)$ is the probability density function for a random variable, for example a normal distribution's PDF is:
$$ g(x) = frac1 {2 pi sigma^2} expleft(frac{- (x - mu)^2 }{ 2 sigma ^2}right) $$
I understand the integral of a PDF gives me the CDF. So:
$$int_{-infty}^0 g(x) , dx$$
Gives me the probability of $x$ being less than $0$. However, what happens when you multiply $g(x)$ by another function $f(x)$ and take the integral? I heard it gives you the expected payoff assuming $f(x)$ is a function of payoffs and you take an integral from -infinity to +infinity. This, if true, I conceptually understand. sum of payoffs times the probabilities is the expected value of whatever game you are playing.
I start getting confused when the boundaries of the integral are not $pm infty$. I'm not sure in that case what integral conceptually means. For example:
$$int_{-infty}^0 f(x) g(x) , dx$$
What does that tell me?
probability distributions normal-distribution random-variable expected-value
edited 5 hours ago
Siong Thye Goh
3,0842621
asked 5 hours ago
vt_ogvt_og
141
New contributor
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to understand conceptually what does the following give me or tell me:
$$int f(x) cdot g(x) , dx$$
where $f(x)$ is any continuous function of $x$ and $g(x)$ is the probability density function for a random variable, for example a normal distribution's PDF is:
$$ g(x) = frac1 {2 pi sigma^2} expleft(frac{- (x - mu)^2 }{ 2 sigma ^2}right) $$
I understand the integral of a PDF gives me the CDF. So:
$$int_{-infty}^0 g(x) , dx$$
Gives me the probability of $x$ being less than $0$. However, what happens when you multiply $g(x)$ by another function $f(x)$ and take the integral? I heard it gives you the expected payoff assuming $f(x)$ is a function of payoffs and you take an integral from -infinity to +infinity. This, if true, I conceptually understand. sum of payoffs times the probabilities is the expected value of whatever game you are playing.
I start getting confused when the boundaries of the integral are not $pm infty$. I'm not sure in that case what integral conceptually means. For example:
$$int_{-infty}^0 f(x) g(x) , dx$$
What does that tell me?
probability distributions normal-distribution random-variable expected-value
edited 5 hours ago
Siong Thye Goh
3,0842621
asked 5 hours ago
vt_ogvt_og
141
New contributor
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to understand conceptually what does the following give me or tell me:
$$int f(x) cdot g(x) , dx$$
where $f(x)$ is any continuous function of $x$ and $g(x)$ is the probability density function for a random variable, for example a normal distribution's PDF is:
$$ g(x) = frac1 {2 pi sigma^2} expleft(frac{- (x - mu)^2 }{ 2 sigma ^2}right) $$
I understand the integral of a PDF gives me the CDF. So:
$$int_{-infty}^0 g(x) , dx$$
Gives me the probability of $x$ being less than $0$. However, what happens when you multiply $g(x)$ by another function $f(x)$ and take the integral? I heard it gives you the expected payoff assuming $f(x)$ is a function of payoffs and you take an integral from -infinity to +infinity. This, if true, I conceptually understand. sum of payoffs times the probabilities is the expected value of whatever game you are playing.
I start getting confused when the boundaries of the integral are not $pm infty$. I'm not sure in that case what integral conceptually means. For example:
$$int_{-infty}^0 f(x) g(x) , dx$$
What does that tell me?
probability distributions normal-distribution random-variable expected-value
probability distributions normal-distribution random-variable expected-value
edited 5 hours ago
Siong Thye Goh
3,0842621
asked 5 hours ago
vt_ogvt_og
141
New contributor
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Siong Thye Goh
3,0842621
asked 5 hours ago
vt_ogvt_og
141
New contributor
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Siong Thye Goh
3,0842621
edited 5 hours ago
Siong Thye Goh
3,0842621
edited 5 hours ago
Siong Thye Goh
3,0842621
3,0842621
asked 5 hours ago
vt_ogvt_og
141
New contributor
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
vt_ogvt_og
141
asked 5 hours ago
vt_ogvt_og
141
141
New contributor
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
vt_og is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Suppose $g$ is the pdf of random variable $X$, then
$$E[f(X)|X in A]= frac{int_A f(x)g(x) , dx}{int_A g(t) , dt}$$
Hence $$int_A f(x) g(x) , dt = Pr(X in A) E[f(X)|X in A],$$
it gives you the product of the conditional expectation of $f(X)$ given that $X in A$ and the probability that $X$ is in $A$.
I think $E[f(X)|X in A]$ is a more interesting quantity, and I would divide your integral with $Pr(X in A)$ to recover it.
answered 5 hours ago
Siong Thye GohSiong Thye Goh
3,0842621
$endgroup$
add a comment |
$begingroup$
The expected value of $X$ following distribution $g$ is
$$
E[X] = int x ,g(x) ,dx
$$
By the law of the unconscious statistician we know that the expected value of a function $f$ of random variable $X$ is
$$
E[f(X)] = int f(x) ,g(x) ,dx
$$
What does it tell you? It tells you what would be the expected value of your random variable if you transformed it somehow. For example, say that you have know what is the distribution of height for humans in centimetres, but are interested of distribution in meters, i.e. $f(x) = x/100$.
answered 4 hours ago
Tim♦Tim
60.8k9134230
$endgroup$
add a comment |
$begingroup$
Another way to look at this integral is through the random variable transformation point of view. You can think of $Y=f(x)$ as a new random variable, and your integral is the expectation (mean) $E[Y]$ of the new variable $Y$.
Let's build the probability density function of the new variable $Y$. Unfortunately, we can't derive an analytical expression. We'll have to use a more complex, integral form of it.
What is the probability that $Y$ will have values between $y$ and $y+dy$? We look at all $x$ where $y<f(x)<y+dy$ and add up their probabilities:
$$g_y(y)dy=int_x mathbb1_{y<f(x)<y+dy} g(x)dx $$
Next, we simply integrate over all values of $Y$:
$$E[y]=int_yyg_y(y)dy$$
Since $int_y$ runs through all possible $y$, it's the same as running the integral through all possible $x$.
Just pause for a moment and agree with me...
Now that you agreed with me you'll see that:
$$E[Y]=int_xf(x)g(x)dx$$
answered 5 hours ago
AksakalAksakal
39.5k452120
$endgroup$
1
$begingroup$
This could use some elaboration. Perhaps an example would help. If you would also bring into consideration the idea of truncation or conditioning (which seems to be the main stumbling block), I think your point would be an excellent answer.
$endgroup$
– whuber♦
4 hours ago
1
$begingroup$
@whuber, i did impossible: explained Lebesque integral without measure theory!
$endgroup$
– Aksakal
4 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $g$ is the pdf of random variable $X$, then
$$E[f(X)|X in A]= frac{int_A f(x)g(x) , dx}{int_A g(t) , dt}$$
Hence $$int_A f(x) g(x) , dt = Pr(X in A) E[f(X)|X in A],$$
it gives you the product of the conditional expectation of $f(X)$ given that $X in A$ and the probability that $X$ is in $A$.
I think $E[f(X)|X in A]$ is a more interesting quantity, and I would divide your integral with $Pr(X in A)$ to recover it.
answered 5 hours ago
Siong Thye GohSiong Thye Goh
3,0842621
$endgroup$
add a comment |
$begingroup$
Suppose $g$ is the pdf of random variable $X$, then
$$E[f(X)|X in A]= frac{int_A f(x)g(x) , dx}{int_A g(t) , dt}$$
Hence $$int_A f(x) g(x) , dt = Pr(X in A) E[f(X)|X in A],$$
it gives you the product of the conditional expectation of $f(X)$ given that $X in A$ and the probability that $X$ is in $A$.
I think $E[f(X)|X in A]$ is a more interesting quantity, and I would divide your integral with $Pr(X in A)$ to recover it.
answered 5 hours ago
Siong Thye GohSiong Thye Goh
3,0842621
$endgroup$
add a comment |
$begingroup$
Suppose $g$ is the pdf of random variable $X$, then
$$E[f(X)|X in A]= frac{int_A f(x)g(x) , dx}{int_A g(t) , dt}$$
Hence $$int_A f(x) g(x) , dt = Pr(X in A) E[f(X)|X in A],$$
it gives you the product of the conditional expectation of $f(X)$ given that $X in A$ and the probability that $X$ is in $A$.
I think $E[f(X)|X in A]$ is a more interesting quantity, and I would divide your integral with $Pr(X in A)$ to recover it.
answered 5 hours ago
Siong Thye GohSiong Thye Goh
3,0842621
$endgroup$
Suppose $g$ is the pdf of random variable $X$, then
$$E[f(X)|X in A]= frac{int_A f(x)g(x) , dx}{int_A g(t) , dt}$$
Hence $$int_A f(x) g(x) , dt = Pr(X in A) E[f(X)|X in A],$$
it gives you the product of the conditional expectation of $f(X)$ given that $X in A$ and the probability that $X$ is in $A$.
I think $E[f(X)|X in A]$ is a more interesting quantity, and I would divide your integral with $Pr(X in A)$ to recover it.
answered 5 hours ago
Siong Thye GohSiong Thye Goh
3,0842621
answered 5 hours ago
Siong Thye GohSiong Thye Goh
3,0842621
answered 5 hours ago
Siong Thye GohSiong Thye Goh
3,0842621
answered 5 hours ago
Siong Thye GohSiong Thye Goh
3,0842621
3,0842621
add a comment |
add a comment |
$begingroup$
The expected value of $X$ following distribution $g$ is
$$
E[X] = int x ,g(x) ,dx
$$
By the law of the unconscious statistician we know that the expected value of a function $f$ of random variable $X$ is
$$
E[f(X)] = int f(x) ,g(x) ,dx
$$
What does it tell you? It tells you what would be the expected value of your random variable if you transformed it somehow. For example, say that you have know what is the distribution of height for humans in centimetres, but are interested of distribution in meters, i.e. $f(x) = x/100$.
answered 4 hours ago
Tim♦Tim
60.8k9134230
$endgroup$
add a comment |
$begingroup$
The expected value of $X$ following distribution $g$ is
$$
E[X] = int x ,g(x) ,dx
$$
By the law of the unconscious statistician we know that the expected value of a function $f$ of random variable $X$ is
$$
E[f(X)] = int f(x) ,g(x) ,dx
$$
What does it tell you? It tells you what would be the expected value of your random variable if you transformed it somehow. For example, say that you have know what is the distribution of height for humans in centimetres, but are interested of distribution in meters, i.e. $f(x) = x/100$.
answered 4 hours ago
Tim♦Tim
60.8k9134230
$endgroup$
add a comment |
$begingroup$
The expected value of $X$ following distribution $g$ is
$$
E[X] = int x ,g(x) ,dx
$$
By the law of the unconscious statistician we know that the expected value of a function $f$ of random variable $X$ is
$$
E[f(X)] = int f(x) ,g(x) ,dx
$$
What does it tell you? It tells you what would be the expected value of your random variable if you transformed it somehow. For example, say that you have know what is the distribution of height for humans in centimetres, but are interested of distribution in meters, i.e. $f(x) = x/100$.
answered 4 hours ago
Tim♦Tim
60.8k9134230
$endgroup$
The expected value of $X$ following distribution $g$ is
$$
E[X] = int x ,g(x) ,dx
$$
By the law of the unconscious statistician we know that the expected value of a function $f$ of random variable $X$ is
$$
E[f(X)] = int f(x) ,g(x) ,dx
$$
What does it tell you? It tells you what would be the expected value of your random variable if you transformed it somehow. For example, say that you have know what is the distribution of height for humans in centimetres, but are interested of distribution in meters, i.e. $f(x) = x/100$.
answered 4 hours ago
Tim♦Tim
60.8k9134230
answered 4 hours ago
Tim♦Tim
60.8k9134230
answered 4 hours ago
Tim♦Tim
60.8k9134230
answered 4 hours ago
Tim♦Tim
60.8k9134230
60.8k9134230
add a comment |
add a comment |
$begingroup$
Another way to look at this integral is through the random variable transformation point of view. You can think of $Y=f(x)$ as a new random variable, and your integral is the expectation (mean) $E[Y]$ of the new variable $Y$.
Let's build the probability density function of the new variable $Y$. Unfortunately, we can't derive an analytical expression. We'll have to use a more complex, integral form of it.
What is the probability that $Y$ will have values between $y$ and $y+dy$? We look at all $x$ where $y<f(x)<y+dy$ and add up their probabilities:
$$g_y(y)dy=int_x mathbb1_{y<f(x)<y+dy} g(x)dx $$
Next, we simply integrate over all values of $Y$:
$$E[y]=int_yyg_y(y)dy$$
Since $int_y$ runs through all possible $y$, it's the same as running the integral through all possible $x$.
Just pause for a moment and agree with me...
Now that you agreed with me you'll see that:
$$E[Y]=int_xf(x)g(x)dx$$
answered 5 hours ago
AksakalAksakal
39.5k452120
$endgroup$
1
$begingroup$
This could use some elaboration. Perhaps an example would help. If you would also bring into consideration the idea of truncation or conditioning (which seems to be the main stumbling block), I think your point would be an excellent answer.
$endgroup$
– whuber♦
4 hours ago
1
$begingroup$
@whuber, i did impossible: explained Lebesque integral without measure theory!
$endgroup$
– Aksakal
4 hours ago
add a comment |
$begingroup$
Another way to look at this integral is through the random variable transformation point of view. You can think of $Y=f(x)$ as a new random variable, and your integral is the expectation (mean) $E[Y]$ of the new variable $Y$.
Let's build the probability density function of the new variable $Y$. Unfortunately, we can't derive an analytical expression. We'll have to use a more complex, integral form of it.
What is the probability that $Y$ will have values between $y$ and $y+dy$? We look at all $x$ where $y<f(x)<y+dy$ and add up their probabilities:
$$g_y(y)dy=int_x mathbb1_{y<f(x)<y+dy} g(x)dx $$
Next, we simply integrate over all values of $Y$:
$$E[y]=int_yyg_y(y)dy$$
Since $int_y$ runs through all possible $y$, it's the same as running the integral through all possible $x$.
Just pause for a moment and agree with me...
Now that you agreed with me you'll see that:
$$E[Y]=int_xf(x)g(x)dx$$
answered 5 hours ago
AksakalAksakal
39.5k452120
$endgroup$
1
$begingroup$
This could use some elaboration. Perhaps an example would help. If you would also bring into consideration the idea of truncation or conditioning (which seems to be the main stumbling block), I think your point would be an excellent answer.
$endgroup$
– whuber♦
4 hours ago
1
$begingroup$
@whuber, i did impossible: explained Lebesque integral without measure theory!
$endgroup$
– Aksakal
4 hours ago
add a comment |
$begingroup$
Another way to look at this integral is through the random variable transformation point of view. You can think of $Y=f(x)$ as a new random variable, and your integral is the expectation (mean) $E[Y]$ of the new variable $Y$.
Let's build the probability density function of the new variable $Y$. Unfortunately, we can't derive an analytical expression. We'll have to use a more complex, integral form of it.
What is the probability that $Y$ will have values between $y$ and $y+dy$? We look at all $x$ where $y<f(x)<y+dy$ and add up their probabilities:
$$g_y(y)dy=int_x mathbb1_{y<f(x)<y+dy} g(x)dx $$
Next, we simply integrate over all values of $Y$:
$$E[y]=int_yyg_y(y)dy$$
Since $int_y$ runs through all possible $y$, it's the same as running the integral through all possible $x$.
Just pause for a moment and agree with me...
Now that you agreed with me you'll see that:
$$E[Y]=int_xf(x)g(x)dx$$
answered 5 hours ago
AksakalAksakal
39.5k452120
$endgroup$
Another way to look at this integral is through the random variable transformation point of view. You can think of $Y=f(x)$ as a new random variable, and your integral is the expectation (mean) $E[Y]$ of the new variable $Y$.
Let's build the probability density function of the new variable $Y$. Unfortunately, we can't derive an analytical expression. We'll have to use a more complex, integral form of it.
What is the probability that $Y$ will have values between $y$ and $y+dy$? We look at all $x$ where $y<f(x)<y+dy$ and add up their probabilities:
$$g_y(y)dy=int_x mathbb1_{y<f(x)<y+dy} g(x)dx $$
Next, we simply integrate over all values of $Y$:
$$E[y]=int_yyg_y(y)dy$$
Since $int_y$ runs through all possible $y$, it's the same as running the integral through all possible $x$.
Just pause for a moment and agree with me...
Now that you agreed with me you'll see that:
$$E[Y]=int_xf(x)g(x)dx$$
answered 5 hours ago
AksakalAksakal
39.5k452120
edited 4 hours ago
answered 5 hours ago
AksakalAksakal
39.5k452120
answered 5 hours ago
AksakalAksakal
39.5k452120
answered 5 hours ago
AksakalAksakal
39.5k452120
39.5k452120
1
$begingroup$
This could use some elaboration. Perhaps an example would help. If you would also bring into consideration the idea of truncation or conditioning (which seems to be the main stumbling block), I think your point would be an excellent answer.
$endgroup$
– whuber♦
4 hours ago
1
$begingroup$
@whuber, i did impossible: explained Lebesque integral without measure theory!
$endgroup$
– Aksakal
4 hours ago
add a comment |
1
$begingroup$
This could use some elaboration. Perhaps an example would help. If you would also bring into consideration the idea of truncation or conditioning (which seems to be the main stumbling block), I think your point would be an excellent answer.
$endgroup$
– whuber♦
4 hours ago
1
$begingroup$
@whuber, i did impossible: explained Lebesque integral without measure theory!
$endgroup$
– Aksakal
4 hours ago
1
1
$begingroup$
This could use some elaboration. Perhaps an example would help. If you would also bring into consideration the idea of truncation or conditioning (which seems to be the main stumbling block), I think your point would be an excellent answer.
$endgroup$
– whuber♦
4 hours ago
$begingroup$
This could use some elaboration. Perhaps an example would help. If you would also bring into consideration the idea of truncation or conditioning (which seems to be the main stumbling block), I think your point would be an excellent answer.
$endgroup$
– whuber♦
4 hours ago
1
1
$begingroup$
@whuber, i did impossible: explained Lebesque integral without measure theory!
$endgroup$
– Aksakal
4 hours ago
$begingroup$
@whuber, i did impossible: explained Lebesque integral without measure theory!
$endgroup$
– Aksakal
4 hours ago
add a comment |
vt_og is a new contributor. Be nice, and check out our Code of Conduct.
vt_og is a new contributor. Be nice, and check out our Code of Conduct.
vt_og is a new contributor. Be nice, and check out our Code of Conduct.
vt_og is a new contributor. Be nice, and check out our Code of Conduct.
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