$Y=u(X)$ be a continuous, decreasing function of $X$ with inverse function $X=v(Y)$, why this equality hold...
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Let $X$ be a continuous random variable with a generic p.d.f. $f(x)$ defined over the support $c_1<x<c_2$. And, let $Y = u(X)$ be a continuous, decreasing function of $X$ with inverse function $X = v(Y)$.
question starts here,
for the portion of the function for which $u(X)leq y$, it is also true that $Xgeq v(Y)$
$p(Y<y)= c= p(X>v(y)) = 1- p(X<v(y))$
plz tell me why $p(Y<y)= p(u(x)<y)$ changed into the $p(X>v(y)) = 1- p(X<v(y))$ what is the logic ??? plz explain with dense information
probability probability-theory probability-distributions conditional-probability
edited Nov 21 at 16:54
gd1035
410210
asked Nov 21 at 16:27
rizwan niaz
13
add a comment |
up vote
-1
down vote
favorite
Let $X$ be a continuous random variable with a generic p.d.f. $f(x)$ defined over the support $c_1<x<c_2$. And, let $Y = u(X)$ be a continuous, decreasing function of $X$ with inverse function $X = v(Y)$.
question starts here,
for the portion of the function for which $u(X)leq y$, it is also true that $Xgeq v(Y)$
$p(Y<y)= c= p(X>v(y)) = 1- p(X<v(y))$
plz tell me why $p(Y<y)= p(u(x)<y)$ changed into the $p(X>v(y)) = 1- p(X<v(y))$ what is the logic ??? plz explain with dense information
probability probability-theory probability-distributions conditional-probability
edited Nov 21 at 16:54
gd1035
410210
asked Nov 21 at 16:27
rizwan niaz
13
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $X$ be a continuous random variable with a generic p.d.f. $f(x)$ defined over the support $c_1<x<c_2$. And, let $Y = u(X)$ be a continuous, decreasing function of $X$ with inverse function $X = v(Y)$.
question starts here,
for the portion of the function for which $u(X)leq y$, it is also true that $Xgeq v(Y)$
$p(Y<y)= c= p(X>v(y)) = 1- p(X<v(y))$
plz tell me why $p(Y<y)= p(u(x)<y)$ changed into the $p(X>v(y)) = 1- p(X<v(y))$ what is the logic ??? plz explain with dense information
probability probability-theory probability-distributions conditional-probability
edited Nov 21 at 16:54
gd1035
410210
asked Nov 21 at 16:27
rizwan niaz
13
Let $X$ be a continuous random variable with a generic p.d.f. $f(x)$ defined over the support $c_1<x<c_2$. And, let $Y = u(X)$ be a continuous, decreasing function of $X$ with inverse function $X = v(Y)$.
question starts here,
for the portion of the function for which $u(X)leq y$, it is also true that $Xgeq v(Y)$
$p(Y<y)= c= p(X>v(y)) = 1- p(X<v(y))$
plz tell me why $p(Y<y)= p(u(x)<y)$ changed into the $p(X>v(y)) = 1- p(X<v(y))$ what is the logic ??? plz explain with dense information
probability probability-theory probability-distributions conditional-probability
probability probability-theory probability-distributions conditional-probability
edited Nov 21 at 16:54
gd1035
410210
asked Nov 21 at 16:27
rizwan niaz
13
edited Nov 21 at 16:54
gd1035
410210
asked Nov 21 at 16:27
rizwan niaz
13
edited Nov 21 at 16:54
gd1035
410210
edited Nov 21 at 16:54
gd1035
410210
edited Nov 21 at 16:54
gd1035
410210
410210
asked Nov 21 at 16:27
rizwan niaz
13
asked Nov 21 at 16:27
rizwan niaz
13
asked Nov 21 at 16:27
rizwan niaz
13
13
add a comment |
add a comment |
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