CDF after Transformation.
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A random variable $Y$ has lognormal ($mu,sigma$) distribution if its probability density function is $$ f(y)=frac{1}{ysigmasqrt{2pi}}exp-frac{(ln y-mu)^2}{2sigma^2}$$ its CDF will be $$hat F(y)=Phileft(frac{log(y)-mu}{sigma}right)$$ If transformation is made as $mu rightarrow log(x)$ and $sigma rightarrow sqrt{4log(1+h)}$ where $h$ is the bandwidth and $x$ is gird then what will be CDF for $$ f(y)=frac{1}{ysqrt{8log(1+h)pi}}exp[-frac{(ln y-ln x)^2}{8ln (1+h)}]$$
Either it is same as usual with transformation like $$hat F(x)= Phileft(frac{log(y)-log(x)}{sqrt{4log(1+h)}}right)$$
or I need to again integrate it with respect to $x$ ? Please share your precious views or solution. Can someone kindly integrate 2nd $f(y)$ with respect to $x$ for limit $(0,x)$
integration statistics probability-distributions transformation
asked Jan 21 at 12:59
AngelAngel
169
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add a comment |
$begingroup$
A random variable $Y$ has lognormal ($mu,sigma$) distribution if its probability density function is $$ f(y)=frac{1}{ysigmasqrt{2pi}}exp-frac{(ln y-mu)^2}{2sigma^2}$$ its CDF will be $$hat F(y)=Phileft(frac{log(y)-mu}{sigma}right)$$ If transformation is made as $mu rightarrow log(x)$ and $sigma rightarrow sqrt{4log(1+h)}$ where $h$ is the bandwidth and $x$ is gird then what will be CDF for $$ f(y)=frac{1}{ysqrt{8log(1+h)pi}}exp[-frac{(ln y-ln x)^2}{8ln (1+h)}]$$
Either it is same as usual with transformation like $$hat F(x)= Phileft(frac{log(y)-log(x)}{sqrt{4log(1+h)}}right)$$
or I need to again integrate it with respect to $x$ ? Please share your precious views or solution. Can someone kindly integrate 2nd $f(y)$ with respect to $x$ for limit $(0,x)$
integration statistics probability-distributions transformation
asked Jan 21 at 12:59
AngelAngel
169
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1
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@Did is this problem of my text of question or way of CDF?
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– Angel
Jan 21 at 13:12
1
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@Did Kindly see my edit.
$endgroup$
– Angel
Jan 21 at 13:42
1
$begingroup$
If you mean (twice) $F(y)$ instead of $hat F(x)$ and that $x$ and $h$ are some fixed parameters then the answer is: Obviously, yes.
$endgroup$
– Did
Jan 21 at 14:16
1
$begingroup$
no, x is not fixed, It is a vector.
$endgroup$
– Angel
Jan 22 at 7:12
1
$begingroup$
I tried as it is to plot CDF in a software, but graph of CDF is decreasing.
$endgroup$
– Angel
Jan 22 at 7:13
add a comment |
$begingroup$
A random variable $Y$ has lognormal ($mu,sigma$) distribution if its probability density function is $$ f(y)=frac{1}{ysigmasqrt{2pi}}exp-frac{(ln y-mu)^2}{2sigma^2}$$ its CDF will be $$hat F(y)=Phileft(frac{log(y)-mu}{sigma}right)$$ If transformation is made as $mu rightarrow log(x)$ and $sigma rightarrow sqrt{4log(1+h)}$ where $h$ is the bandwidth and $x$ is gird then what will be CDF for $$ f(y)=frac{1}{ysqrt{8log(1+h)pi}}exp[-frac{(ln y-ln x)^2}{8ln (1+h)}]$$
Either it is same as usual with transformation like $$hat F(x)= Phileft(frac{log(y)-log(x)}{sqrt{4log(1+h)}}right)$$
or I need to again integrate it with respect to $x$ ? Please share your precious views or solution. Can someone kindly integrate 2nd $f(y)$ with respect to $x$ for limit $(0,x)$
integration statistics probability-distributions transformation
asked Jan 21 at 12:59
AngelAngel
169
$endgroup$
A random variable $Y$ has lognormal ($mu,sigma$) distribution if its probability density function is $$ f(y)=frac{1}{ysigmasqrt{2pi}}exp-frac{(ln y-mu)^2}{2sigma^2}$$ its CDF will be $$hat F(y)=Phileft(frac{log(y)-mu}{sigma}right)$$ If transformation is made as $mu rightarrow log(x)$ and $sigma rightarrow sqrt{4log(1+h)}$ where $h$ is the bandwidth and $x$ is gird then what will be CDF for $$ f(y)=frac{1}{ysqrt{8log(1+h)pi}}exp[-frac{(ln y-ln x)^2}{8ln (1+h)}]$$
Either it is same as usual with transformation like $$hat F(x)= Phileft(frac{log(y)-log(x)}{sqrt{4log(1+h)}}right)$$
or I need to again integrate it with respect to $x$ ? Please share your precious views or solution. Can someone kindly integrate 2nd $f(y)$ with respect to $x$ for limit $(0,x)$
integration statistics probability-distributions transformation
integration statistics probability-distributions transformation
asked Jan 21 at 12:59
AngelAngel
169
asked Jan 21 at 12:59
AngelAngel
169
edited Jan 22 at 11:09
Angel
asked Jan 21 at 12:59
AngelAngel
169
asked Jan 21 at 12:59
AngelAngel
169
asked Jan 21 at 12:59
AngelAngel
169
169
1
$begingroup$
@Did is this problem of my text of question or way of CDF?
$endgroup$
– Angel
Jan 21 at 13:12
1
$begingroup$
@Did Kindly see my edit.
$endgroup$
– Angel
Jan 21 at 13:42
1
$begingroup$
If you mean (twice) $F(y)$ instead of $hat F(x)$ and that $x$ and $h$ are some fixed parameters then the answer is: Obviously, yes.
$endgroup$
– Did
Jan 21 at 14:16
1
$begingroup$
no, x is not fixed, It is a vector.
$endgroup$
– Angel
Jan 22 at 7:12
1
$begingroup$
I tried as it is to plot CDF in a software, but graph of CDF is decreasing.
$endgroup$
– Angel
Jan 22 at 7:13
add a comment |
1
$begingroup$
@Did is this problem of my text of question or way of CDF?
$endgroup$
– Angel
Jan 21 at 13:12
1
$begingroup$
@Did Kindly see my edit.
$endgroup$
– Angel
Jan 21 at 13:42
1
$begingroup$
If you mean (twice) $F(y)$ instead of $hat F(x)$ and that $x$ and $h$ are some fixed parameters then the answer is: Obviously, yes.
$endgroup$
– Did
Jan 21 at 14:16
1
$begingroup$
no, x is not fixed, It is a vector.
$endgroup$
– Angel
Jan 22 at 7:12
1
$begingroup$
I tried as it is to plot CDF in a software, but graph of CDF is decreasing.
$endgroup$
– Angel
Jan 22 at 7:13
1
1
$begingroup$
@Did is this problem of my text of question or way of CDF?
$endgroup$
– Angel
Jan 21 at 13:12
$begingroup$
@Did is this problem of my text of question or way of CDF?
$endgroup$
– Angel
Jan 21 at 13:12
1
1
$begingroup$
@Did Kindly see my edit.
$endgroup$
– Angel
Jan 21 at 13:42
$begingroup$
@Did Kindly see my edit.
$endgroup$
– Angel
Jan 21 at 13:42
1
1
$begingroup$
If you mean (twice) $F(y)$ instead of $hat F(x)$ and that $x$ and $h$ are some fixed parameters then the answer is: Obviously, yes.
$endgroup$
– Did
Jan 21 at 14:16
$begingroup$
If you mean (twice) $F(y)$ instead of $hat F(x)$ and that $x$ and $h$ are some fixed parameters then the answer is: Obviously, yes.
$endgroup$
– Did
Jan 21 at 14:16
1
1
$begingroup$
no, x is not fixed, It is a vector.
$endgroup$
– Angel
Jan 22 at 7:12
$begingroup$
no, x is not fixed, It is a vector.
$endgroup$
– Angel
Jan 22 at 7:12
1
1
$begingroup$
I tried as it is to plot CDF in a software, but graph of CDF is decreasing.
$endgroup$
– Angel
Jan 22 at 7:13
$begingroup$
I tried as it is to plot CDF in a software, but graph of CDF is decreasing.
$endgroup$
– Angel
Jan 22 at 7:13
add a comment |
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1
$begingroup$
@Did is this problem of my text of question or way of CDF?
$endgroup$
– Angel
Jan 21 at 13:12
1
$begingroup$
@Did Kindly see my edit.
$endgroup$
– Angel
Jan 21 at 13:42
1
$begingroup$
If you mean (twice) $F(y)$ instead of $hat F(x)$ and that $x$ and $h$ are some fixed parameters then the answer is: Obviously, yes.
$endgroup$
– Did
Jan 21 at 14:16
1
$begingroup$
no, x is not fixed, It is a vector.
$endgroup$
– Angel
Jan 22 at 7:12
1
$begingroup$
I tried as it is to plot CDF in a software, but graph of CDF is decreasing.
$endgroup$
– Angel
Jan 22 at 7:13