Intuitive interpretation for some variations of random variable times its pdf
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I need help to interpret the following expression.
D,w = random variable
f(D) = probability distribution function
$A = int_0^infty f(D) D dD$
then, A is the mean for all D values that we have, this is the definition of expected value.
$B = int_w^infty f(D) (D-w) dD$
Does it mean that B is the mean for all D values which are larger than w?
$G = int_w^infty f(D) frac{(D-w)}{w}dD$
This one, I really don't get it. What does it mean?
Can someone give me the intuitive understanding for those 2 expression (B and G)?
This picture is from the paper that I am trying to understand.
statistics
asked Jan 8 at 10:19
Codelearner777Codelearner777
427
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add a comment |
$begingroup$
I need help to interpret the following expression.
D,w = random variable
f(D) = probability distribution function
$A = int_0^infty f(D) D dD$
then, A is the mean for all D values that we have, this is the definition of expected value.
$B = int_w^infty f(D) (D-w) dD$
Does it mean that B is the mean for all D values which are larger than w?
$G = int_w^infty f(D) frac{(D-w)}{w}dD$
This one, I really don't get it. What does it mean?
Can someone give me the intuitive understanding for those 2 expression (B and G)?
This picture is from the paper that I am trying to understand.
statistics
asked Jan 8 at 10:19
Codelearner777Codelearner777
427
$endgroup$
$begingroup$
Presumably $dfrac{D-w}{w}$ is supposed to be the probability that a large grain spans the linewidth
$endgroup$
– Henry
Jan 8 at 10:26
$begingroup$
It's hard to help without more context. Your $A$ is indeed $E[D]$, but B and G do not seem to have some direct interpretation (without context).
$endgroup$
– leonbloy
Jan 8 at 15:39
add a comment |
$begingroup$
I need help to interpret the following expression.
D,w = random variable
f(D) = probability distribution function
$A = int_0^infty f(D) D dD$
then, A is the mean for all D values that we have, this is the definition of expected value.
$B = int_w^infty f(D) (D-w) dD$
Does it mean that B is the mean for all D values which are larger than w?
$G = int_w^infty f(D) frac{(D-w)}{w}dD$
This one, I really don't get it. What does it mean?
Can someone give me the intuitive understanding for those 2 expression (B and G)?
This picture is from the paper that I am trying to understand.
statistics
asked Jan 8 at 10:19
Codelearner777Codelearner777
427
$endgroup$
I need help to interpret the following expression.
D,w = random variable
f(D) = probability distribution function
$A = int_0^infty f(D) D dD$
then, A is the mean for all D values that we have, this is the definition of expected value.
$B = int_w^infty f(D) (D-w) dD$
Does it mean that B is the mean for all D values which are larger than w?
$G = int_w^infty f(D) frac{(D-w)}{w}dD$
This one, I really don't get it. What does it mean?
Can someone give me the intuitive understanding for those 2 expression (B and G)?
This picture is from the paper that I am trying to understand.
statistics
statistics
asked Jan 8 at 10:19
Codelearner777Codelearner777
427
asked Jan 8 at 10:19
Codelearner777Codelearner777
427
edited Jan 8 at 15:18
Codelearner777
asked Jan 8 at 10:19
Codelearner777Codelearner777
427
asked Jan 8 at 10:19
Codelearner777Codelearner777
427
asked Jan 8 at 10:19
Codelearner777Codelearner777
427
427
$begingroup$
Presumably $dfrac{D-w}{w}$ is supposed to be the probability that a large grain spans the linewidth
$endgroup$
– Henry
Jan 8 at 10:26
$begingroup$
It's hard to help without more context. Your $A$ is indeed $E[D]$, but B and G do not seem to have some direct interpretation (without context).
$endgroup$
– leonbloy
Jan 8 at 15:39
add a comment |
$begingroup$
Presumably $dfrac{D-w}{w}$ is supposed to be the probability that a large grain spans the linewidth
$endgroup$
– Henry
Jan 8 at 10:26
$begingroup$
It's hard to help without more context. Your $A$ is indeed $E[D]$, but B and G do not seem to have some direct interpretation (without context).
$endgroup$
– leonbloy
Jan 8 at 15:39
$begingroup$
Presumably $dfrac{D-w}{w}$ is supposed to be the probability that a large grain spans the linewidth
$endgroup$
– Henry
Jan 8 at 10:26
$begingroup$
Presumably $dfrac{D-w}{w}$ is supposed to be the probability that a large grain spans the linewidth
$endgroup$
– Henry
Jan 8 at 10:26
$begingroup$
It's hard to help without more context. Your $A$ is indeed $E[D]$, but B and G do not seem to have some direct interpretation (without context).
$endgroup$
– leonbloy
Jan 8 at 15:39
$begingroup$
It's hard to help without more context. Your $A$ is indeed $E[D]$, but B and G do not seem to have some direct interpretation (without context).
$endgroup$
– leonbloy
Jan 8 at 15:39
add a comment |
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$begingroup$
Presumably $dfrac{D-w}{w}$ is supposed to be the probability that a large grain spans the linewidth
$endgroup$
– Henry
Jan 8 at 10:26
$begingroup$
It's hard to help without more context. Your $A$ is indeed $E[D]$, but B and G do not seem to have some direct interpretation (without context).
$endgroup$
– leonbloy
Jan 8 at 15:39