CDF evaluated at random variable
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Suppose that $Ysim F_Y$ is continuous. Then, we have: $P(Yleq y)=F_Y(y)$. We also have that $F_Y(Y)sim U_{[0,1]}$. But at the same time, $F_Y(Y)=P(Yleq Y)=1$.
So it seems to me that $F_Y(Y)=P(Yleq Y)neq F_Ycirc Ysim U_{[0,1]}$.
Is there a contradiction in my argument?
probability probability-distributions
asked Nov 20 at 15:59
julian.marr
349112
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Suppose that $Ysim F_Y$ is continuous. Then, we have: $P(Yleq y)=F_Y(y)$. We also have that $F_Y(Y)sim U_{[0,1]}$. But at the same time, $F_Y(Y)=P(Yleq Y)=1$.
So it seems to me that $F_Y(Y)=P(Yleq Y)neq F_Ycirc Ysim U_{[0,1]}$.
Is there a contradiction in my argument?
probability probability-distributions
asked Nov 20 at 15:59
julian.marr
349112
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $Ysim F_Y$ is continuous. Then, we have: $P(Yleq y)=F_Y(y)$. We also have that $F_Y(Y)sim U_{[0,1]}$. But at the same time, $F_Y(Y)=P(Yleq Y)=1$.
So it seems to me that $F_Y(Y)=P(Yleq Y)neq F_Ycirc Ysim U_{[0,1]}$.
Is there a contradiction in my argument?
probability probability-distributions
asked Nov 20 at 15:59
julian.marr
349112
Suppose that $Ysim F_Y$ is continuous. Then, we have: $P(Yleq y)=F_Y(y)$. We also have that $F_Y(Y)sim U_{[0,1]}$. But at the same time, $F_Y(Y)=P(Yleq Y)=1$.
So it seems to me that $F_Y(Y)=P(Yleq Y)neq F_Ycirc Ysim U_{[0,1]}$.
Is there a contradiction in my argument?
probability probability-distributions
probability probability-distributions
asked Nov 20 at 15:59
julian.marr
349112
asked Nov 20 at 15:59
julian.marr
349112
asked Nov 20 at 15:59
julian.marr
349112
asked Nov 20 at 15:59
julian.marr
349112
asked Nov 20 at 15:59
julian.marr
349112
349112
add a comment |
add a comment |
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