Show how to form a location-scale family $g(x)$ based on $f(x)$ such that $g(x)$ has mean $0$ and variance...
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Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}e^{{−0.5 (logx)^2}}$, $x >0$.
Show how to form a location-scale family $g(x)$ based on $f(x)$ such that $g(x)$ has mean $0$ and variance $1$.
I'm not sure how to go about this. I know that a location-scale family is a family of probability distributions parametrized by a location parameter and a non-negative scale parameter.
probability-theory statistics probability-distributions random-variables statistical-inference
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add a comment |
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Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}e^{{−0.5 (logx)^2}}$, $x >0$.
Show how to form a location-scale family $g(x)$ based on $f(x)$ such that $g(x)$ has mean $0$ and variance $1$.
I'm not sure how to go about this. I know that a location-scale family is a family of probability distributions parametrized by a location parameter and a non-negative scale parameter.
probability-theory statistics probability-distributions random-variables statistical-inference
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It might be worth starting by finding the mean and variance of $X$
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– Henry
Jan 31 at 0:31
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$E(X)=sqrt{e}$, $E(X^2)=e^2$, and $Var(X)=e(e-1)$
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– ddswsd
Jan 31 at 2:04
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Then you want $Y=dfrac{X-sqrt{e}}{sqrt{e(e-1)}}$ and you can use the expressions at en.wikipedia.org/wiki/Location%E2%80%93scale_family
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– Henry
Jan 31 at 8:11
add a comment |
$begingroup$
Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}e^{{−0.5 (logx)^2}}$, $x >0$.
Show how to form a location-scale family $g(x)$ based on $f(x)$ such that $g(x)$ has mean $0$ and variance $1$.
I'm not sure how to go about this. I know that a location-scale family is a family of probability distributions parametrized by a location parameter and a non-negative scale parameter.
probability-theory statistics probability-distributions random-variables statistical-inference
$endgroup$
Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}e^{{−0.5 (logx)^2}}$, $x >0$.
Show how to form a location-scale family $g(x)$ based on $f(x)$ such that $g(x)$ has mean $0$ and variance $1$.
I'm not sure how to go about this. I know that a location-scale family is a family of probability distributions parametrized by a location parameter and a non-negative scale parameter.
probability-theory statistics probability-distributions random-variables statistical-inference
probability-theory statistics probability-distributions random-variables statistical-inference
asked Jan 30 at 22:46
ddswsdddswsd
37929
37929
$begingroup$
It might be worth starting by finding the mean and variance of $X$
$endgroup$
– Henry
Jan 31 at 0:31
$begingroup$
$E(X)=sqrt{e}$, $E(X^2)=e^2$, and $Var(X)=e(e-1)$
$endgroup$
– ddswsd
Jan 31 at 2:04
$begingroup$
Then you want $Y=dfrac{X-sqrt{e}}{sqrt{e(e-1)}}$ and you can use the expressions at en.wikipedia.org/wiki/Location%E2%80%93scale_family
$endgroup$
– Henry
Jan 31 at 8:11
add a comment |
$begingroup$
It might be worth starting by finding the mean and variance of $X$
$endgroup$
– Henry
Jan 31 at 0:31
$begingroup$
$E(X)=sqrt{e}$, $E(X^2)=e^2$, and $Var(X)=e(e-1)$
$endgroup$
– ddswsd
Jan 31 at 2:04
$begingroup$
Then you want $Y=dfrac{X-sqrt{e}}{sqrt{e(e-1)}}$ and you can use the expressions at en.wikipedia.org/wiki/Location%E2%80%93scale_family
$endgroup$
– Henry
Jan 31 at 8:11
$begingroup$
It might be worth starting by finding the mean and variance of $X$
$endgroup$
– Henry
Jan 31 at 0:31
$begingroup$
It might be worth starting by finding the mean and variance of $X$
$endgroup$
– Henry
Jan 31 at 0:31
$begingroup$
$E(X)=sqrt{e}$, $E(X^2)=e^2$, and $Var(X)=e(e-1)$
$endgroup$
– ddswsd
Jan 31 at 2:04
$begingroup$
$E(X)=sqrt{e}$, $E(X^2)=e^2$, and $Var(X)=e(e-1)$
$endgroup$
– ddswsd
Jan 31 at 2:04
$begingroup$
Then you want $Y=dfrac{X-sqrt{e}}{sqrt{e(e-1)}}$ and you can use the expressions at en.wikipedia.org/wiki/Location%E2%80%93scale_family
$endgroup$
– Henry
Jan 31 at 8:11
$begingroup$
Then you want $Y=dfrac{X-sqrt{e}}{sqrt{e(e-1)}}$ and you can use the expressions at en.wikipedia.org/wiki/Location%E2%80%93scale_family
$endgroup$
– Henry
Jan 31 at 8:11
add a comment |
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$begingroup$
It might be worth starting by finding the mean and variance of $X$
$endgroup$
– Henry
Jan 31 at 0:31
$begingroup$
$E(X)=sqrt{e}$, $E(X^2)=e^2$, and $Var(X)=e(e-1)$
$endgroup$
– ddswsd
Jan 31 at 2:04
$begingroup$
Then you want $Y=dfrac{X-sqrt{e}}{sqrt{e(e-1)}}$ and you can use the expressions at en.wikipedia.org/wiki/Location%E2%80%93scale_family
$endgroup$
– Henry
Jan 31 at 8:11