Proof or relation between a Uniform and Exponential
0
$begingroup$
Given $Xsim U(0,1)$, i have to determine the density of $Y=-frac{1}{lambda}lnx$.
I can't apply the law of transformation of random variables because $g(X)$ is not a monotonic function. So, i write:
$F_Y(y)=mathbb{P}(Yleq y)=mathbb{P}(-lnXleq lambda y)=mathbb{P}(Xgeq e^{-lambda y})=1-mathbb{P}(Xleq e^{-lambda y})=1-F_X(e^{-lambda y})$
Now it's clear that $f_Y(y)=lambda e^{-lambda y}$, but I'm having difficulties to formalize the passage between $1-F_X(e^{-lambda y})$ and $f_Y(y)=lambda e^{-lambda y}$. Anyone can help me?
Thanks in advance!
probability random-variables exponential-function uniform-distribution density-function
asked Dec 8 '18 at 11:16
Marco PittellaMarco Pittella
1338
$endgroup$
add a comment |
0
$begingroup$
Given $Xsim U(0,1)$, i have to determine the density of $Y=-frac{1}{lambda}lnx$.
I can't apply the law of transformation of random variables because $g(X)$ is not a monotonic function. So, i write:
$F_Y(y)=mathbb{P}(Yleq y)=mathbb{P}(-lnXleq lambda y)=mathbb{P}(Xgeq e^{-lambda y})=1-mathbb{P}(Xleq e^{-lambda y})=1-F_X(e^{-lambda y})$
Now it's clear that $f_Y(y)=lambda e^{-lambda y}$, but I'm having difficulties to formalize the passage between $1-F_X(e^{-lambda y})$ and $f_Y(y)=lambda e^{-lambda y}$. Anyone can help me?
Thanks in advance!
probability random-variables exponential-function uniform-distribution density-function
asked Dec 8 '18 at 11:16
Marco PittellaMarco Pittella
1338
$endgroup$
$begingroup$
Edit: $g(X)$ is a monotonic function, so I can apply the law of transformation.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:41
add a comment |
0
0
0
$begingroup$
Given $Xsim U(0,1)$, i have to determine the density of $Y=-frac{1}{lambda}lnx$.
I can't apply the law of transformation of random variables because $g(X)$ is not a monotonic function. So, i write:
$F_Y(y)=mathbb{P}(Yleq y)=mathbb{P}(-lnXleq lambda y)=mathbb{P}(Xgeq e^{-lambda y})=1-mathbb{P}(Xleq e^{-lambda y})=1-F_X(e^{-lambda y})$
Now it's clear that $f_Y(y)=lambda e^{-lambda y}$, but I'm having difficulties to formalize the passage between $1-F_X(e^{-lambda y})$ and $f_Y(y)=lambda e^{-lambda y}$. Anyone can help me?
Thanks in advance!
probability random-variables exponential-function uniform-distribution density-function
asked Dec 8 '18 at 11:16
Marco PittellaMarco Pittella
1338
$endgroup$
Given $Xsim U(0,1)$, i have to determine the density of $Y=-frac{1}{lambda}lnx$.
I can't apply the law of transformation of random variables because $g(X)$ is not a monotonic function. So, i write:
$F_Y(y)=mathbb{P}(Yleq y)=mathbb{P}(-lnXleq lambda y)=mathbb{P}(Xgeq e^{-lambda y})=1-mathbb{P}(Xleq e^{-lambda y})=1-F_X(e^{-lambda y})$
Now it's clear that $f_Y(y)=lambda e^{-lambda y}$, but I'm having difficulties to formalize the passage between $1-F_X(e^{-lambda y})$ and $f_Y(y)=lambda e^{-lambda y}$. Anyone can help me?
Thanks in advance!
probability random-variables exponential-function uniform-distribution density-function
probability random-variables exponential-function uniform-distribution density-function
asked Dec 8 '18 at 11:16
Marco PittellaMarco Pittella
1338
asked Dec 8 '18 at 11:16
Marco PittellaMarco Pittella
1338
edited Dec 11 '18 at 12:03
Marco Pittella
asked Dec 8 '18 at 11:16
Marco PittellaMarco Pittella
1338
asked Dec 8 '18 at 11:16
Marco PittellaMarco Pittella
1338
asked Dec 8 '18 at 11:16
Marco PittellaMarco Pittella
1338
1338
$begingroup$
Edit: $g(X)$ is a monotonic function, so I can apply the law of transformation.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:41
add a comment |
$begingroup$
Edit: $g(X)$ is a monotonic function, so I can apply the law of transformation.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:41
$begingroup$
Edit: $g(X)$ is a monotonic function, so I can apply the law of transformation.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:41
$begingroup$
Edit: $g(X)$ is a monotonic function, so I can apply the law of transformation.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:41
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Realize that $Y $ takes positive values and in your calculation of $F_Y (y) $ preassume that $y>0$.
Go one step further in the calculation and write: $$F_Y (y)=1-e^{-lambda y} $$
This is allowed because $F_X (x)=x $ for $xin (0,1) $.
Now take the derivative of $F_Y (y) $ and you are ready.
answered Dec 8 '18 at 11:28
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you very much for your answer. I don't understand why, if $F_X(f)$ is defined for $x in [0,1]$, so $F_Y(y)=1-F_X(e^{-lambda y})=1-e^{-lambda y}$. I know that logarithm is defined only for positive values: in fact, for both $x<0$ and $x=0$ we know that $Ynotin mathbb{R}$. Moreover, if $Xin [0,1]$ e $Y$ is a linear transformation of $X$, so $e^{-lambda y}in [0,1]$. But i don't understand how this help me to formalize the passage above.
$endgroup$
– Marco Pittella
Dec 10 '18 at 7:07
$begingroup$
I fail to see your problem. 1) Do you agree that $F_Y(y)=1-F_X(e^{-lambda y})$ for $y>0$? 2) Do you agree that $e^{-lambda y}in(0,1)$ for $y>0$ and $lambda>0$? 3) Do you agree that $F_X(x)=x$ for $xin(0,1)$? If the answer on all questions is "yes" then what withholds you from drawing the conclusion that $F_Y(y)=1-e^{-lambda y}$ for $y>0$? Further it is evident that $F_Y(y)=0$ for $yleq0$ so the complete CDF of $Y$ has been found.
$endgroup$
– drhab
Dec 10 '18 at 9:19
$begingroup$
1) Why $y>0$? 2) Why $e^{-lambda y}in (0,1)$
$endgroup$
– Marco Pittella
Dec 10 '18 at 15:51
$begingroup$
Point 1): $X$ only takes values in $(0,1)$. Then automatically $Y=-frac1{lambda}ln X$ only takes values in $(0,infty)$. That makes it clear immediately that $F_Y(y)=P(Yleq y)=0$ for $yleq0$.Then to accomplish our effort to find $F_Y(y)$ it remains to find $F_Y(y)$ for $y>0$. Point 2): If $y>0$ then automatically $e^{-lambda y}in(0,1)$. BTW, you haven't answered the questions I posed you in my former comment.
$endgroup$
– drhab
Dec 10 '18 at 18:16
$begingroup$
Thanks again for your answer. I didn't answer your questions because I wanted to make sure I got this straight. So, yes on all questions. Anyway i understood. Using CDF of $X$, which means that $0<e^{-lambda y}leq 1$, i can write $P(Xgeq e^{-lambda y})=int_{e^{-lambda y}}^{infty}f_X(x)dx=int_{e^{-lambda y}}^{1}1dx+int_{1}^{infty}0dx=int_{e^{-lambda y}}^{1}1dx=[x]_{e^{-lambda y}}^{1}=1-e^{-lambda y}$ or $1-P(Xleq e^{-lambda y})=1-int_{-infty}^{e^{-lambda y}}f_X(x)dx=1-[int_{-infty}^{0}0dx+int_{0}^{e^{-lambda y}}1dx]=1-[x]_{0}^{e^{-lambda y}}=1-e^{-lambda y}$.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:39
|
show 3 more comments
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Realize that $Y $ takes positive values and in your calculation of $F_Y (y) $ preassume that $y>0$.
Go one step further in the calculation and write: $$F_Y (y)=1-e^{-lambda y} $$
This is allowed because $F_X (x)=x $ for $xin (0,1) $.
Now take the derivative of $F_Y (y) $ and you are ready.
answered Dec 8 '18 at 11:28
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you very much for your answer. I don't understand why, if $F_X(f)$ is defined for $x in [0,1]$, so $F_Y(y)=1-F_X(e^{-lambda y})=1-e^{-lambda y}$. I know that logarithm is defined only for positive values: in fact, for both $x<0$ and $x=0$ we know that $Ynotin mathbb{R}$. Moreover, if $Xin [0,1]$ e $Y$ is a linear transformation of $X$, so $e^{-lambda y}in [0,1]$. But i don't understand how this help me to formalize the passage above.
$endgroup$
– Marco Pittella
Dec 10 '18 at 7:07
$begingroup$
I fail to see your problem. 1) Do you agree that $F_Y(y)=1-F_X(e^{-lambda y})$ for $y>0$? 2) Do you agree that $e^{-lambda y}in(0,1)$ for $y>0$ and $lambda>0$? 3) Do you agree that $F_X(x)=x$ for $xin(0,1)$? If the answer on all questions is "yes" then what withholds you from drawing the conclusion that $F_Y(y)=1-e^{-lambda y}$ for $y>0$? Further it is evident that $F_Y(y)=0$ for $yleq0$ so the complete CDF of $Y$ has been found.
$endgroup$
– drhab
Dec 10 '18 at 9:19
$begingroup$
1) Why $y>0$? 2) Why $e^{-lambda y}in (0,1)$
$endgroup$
– Marco Pittella
Dec 10 '18 at 15:51
$begingroup$
Point 1): $X$ only takes values in $(0,1)$. Then automatically $Y=-frac1{lambda}ln X$ only takes values in $(0,infty)$. That makes it clear immediately that $F_Y(y)=P(Yleq y)=0$ for $yleq0$.Then to accomplish our effort to find $F_Y(y)$ it remains to find $F_Y(y)$ for $y>0$. Point 2): If $y>0$ then automatically $e^{-lambda y}in(0,1)$. BTW, you haven't answered the questions I posed you in my former comment.
$endgroup$
– drhab
Dec 10 '18 at 18:16
$begingroup$
Thanks again for your answer. I didn't answer your questions because I wanted to make sure I got this straight. So, yes on all questions. Anyway i understood. Using CDF of $X$, which means that $0<e^{-lambda y}leq 1$, i can write $P(Xgeq e^{-lambda y})=int_{e^{-lambda y}}^{infty}f_X(x)dx=int_{e^{-lambda y}}^{1}1dx+int_{1}^{infty}0dx=int_{e^{-lambda y}}^{1}1dx=[x]_{e^{-lambda y}}^{1}=1-e^{-lambda y}$ or $1-P(Xleq e^{-lambda y})=1-int_{-infty}^{e^{-lambda y}}f_X(x)dx=1-[int_{-infty}^{0}0dx+int_{0}^{e^{-lambda y}}1dx]=1-[x]_{0}^{e^{-lambda y}}=1-e^{-lambda y}$.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:39
|
show 3 more comments
1
$begingroup$
Realize that $Y $ takes positive values and in your calculation of $F_Y (y) $ preassume that $y>0$.
Go one step further in the calculation and write: $$F_Y (y)=1-e^{-lambda y} $$
This is allowed because $F_X (x)=x $ for $xin (0,1) $.
Now take the derivative of $F_Y (y) $ and you are ready.
answered Dec 8 '18 at 11:28
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you very much for your answer. I don't understand why, if $F_X(f)$ is defined for $x in [0,1]$, so $F_Y(y)=1-F_X(e^{-lambda y})=1-e^{-lambda y}$. I know that logarithm is defined only for positive values: in fact, for both $x<0$ and $x=0$ we know that $Ynotin mathbb{R}$. Moreover, if $Xin [0,1]$ e $Y$ is a linear transformation of $X$, so $e^{-lambda y}in [0,1]$. But i don't understand how this help me to formalize the passage above.
$endgroup$
– Marco Pittella
Dec 10 '18 at 7:07
$begingroup$
I fail to see your problem. 1) Do you agree that $F_Y(y)=1-F_X(e^{-lambda y})$ for $y>0$? 2) Do you agree that $e^{-lambda y}in(0,1)$ for $y>0$ and $lambda>0$? 3) Do you agree that $F_X(x)=x$ for $xin(0,1)$? If the answer on all questions is "yes" then what withholds you from drawing the conclusion that $F_Y(y)=1-e^{-lambda y}$ for $y>0$? Further it is evident that $F_Y(y)=0$ for $yleq0$ so the complete CDF of $Y$ has been found.
$endgroup$
– drhab
Dec 10 '18 at 9:19
$begingroup$
1) Why $y>0$? 2) Why $e^{-lambda y}in (0,1)$
$endgroup$
– Marco Pittella
Dec 10 '18 at 15:51
$begingroup$
Point 1): $X$ only takes values in $(0,1)$. Then automatically $Y=-frac1{lambda}ln X$ only takes values in $(0,infty)$. That makes it clear immediately that $F_Y(y)=P(Yleq y)=0$ for $yleq0$.Then to accomplish our effort to find $F_Y(y)$ it remains to find $F_Y(y)$ for $y>0$. Point 2): If $y>0$ then automatically $e^{-lambda y}in(0,1)$. BTW, you haven't answered the questions I posed you in my former comment.
$endgroup$
– drhab
Dec 10 '18 at 18:16
$begingroup$
Thanks again for your answer. I didn't answer your questions because I wanted to make sure I got this straight. So, yes on all questions. Anyway i understood. Using CDF of $X$, which means that $0<e^{-lambda y}leq 1$, i can write $P(Xgeq e^{-lambda y})=int_{e^{-lambda y}}^{infty}f_X(x)dx=int_{e^{-lambda y}}^{1}1dx+int_{1}^{infty}0dx=int_{e^{-lambda y}}^{1}1dx=[x]_{e^{-lambda y}}^{1}=1-e^{-lambda y}$ or $1-P(Xleq e^{-lambda y})=1-int_{-infty}^{e^{-lambda y}}f_X(x)dx=1-[int_{-infty}^{0}0dx+int_{0}^{e^{-lambda y}}1dx]=1-[x]_{0}^{e^{-lambda y}}=1-e^{-lambda y}$.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:39
|
show 3 more comments
1
1
1
$begingroup$
Realize that $Y $ takes positive values and in your calculation of $F_Y (y) $ preassume that $y>0$.
Go one step further in the calculation and write: $$F_Y (y)=1-e^{-lambda y} $$
This is allowed because $F_X (x)=x $ for $xin (0,1) $.
Now take the derivative of $F_Y (y) $ and you are ready.
answered Dec 8 '18 at 11:28
drhabdrhab
103k545136
$endgroup$
Realize that $Y $ takes positive values and in your calculation of $F_Y (y) $ preassume that $y>0$.
Go one step further in the calculation and write: $$F_Y (y)=1-e^{-lambda y} $$
This is allowed because $F_X (x)=x $ for $xin (0,1) $.
Now take the derivative of $F_Y (y) $ and you are ready.
answered Dec 8 '18 at 11:28
drhabdrhab
103k545136
edited Dec 8 '18 at 11:33
answered Dec 8 '18 at 11:28
drhabdrhab
103k545136
answered Dec 8 '18 at 11:28
drhabdrhab
103k545136
answered Dec 8 '18 at 11:28
drhabdrhab
103k545136
103k545136
$begingroup$
Thank you very much for your answer. I don't understand why, if $F_X(f)$ is defined for $x in [0,1]$, so $F_Y(y)=1-F_X(e^{-lambda y})=1-e^{-lambda y}$. I know that logarithm is defined only for positive values: in fact, for both $x<0$ and $x=0$ we know that $Ynotin mathbb{R}$. Moreover, if $Xin [0,1]$ e $Y$ is a linear transformation of $X$, so $e^{-lambda y}in [0,1]$. But i don't understand how this help me to formalize the passage above.
$endgroup$
– Marco Pittella
Dec 10 '18 at 7:07
$begingroup$
I fail to see your problem. 1) Do you agree that $F_Y(y)=1-F_X(e^{-lambda y})$ for $y>0$? 2) Do you agree that $e^{-lambda y}in(0,1)$ for $y>0$ and $lambda>0$? 3) Do you agree that $F_X(x)=x$ for $xin(0,1)$? If the answer on all questions is "yes" then what withholds you from drawing the conclusion that $F_Y(y)=1-e^{-lambda y}$ for $y>0$? Further it is evident that $F_Y(y)=0$ for $yleq0$ so the complete CDF of $Y$ has been found.
$endgroup$
– drhab
Dec 10 '18 at 9:19
$begingroup$
1) Why $y>0$? 2) Why $e^{-lambda y}in (0,1)$
$endgroup$
– Marco Pittella
Dec 10 '18 at 15:51
$begingroup$
Point 1): $X$ only takes values in $(0,1)$. Then automatically $Y=-frac1{lambda}ln X$ only takes values in $(0,infty)$. That makes it clear immediately that $F_Y(y)=P(Yleq y)=0$ for $yleq0$.Then to accomplish our effort to find $F_Y(y)$ it remains to find $F_Y(y)$ for $y>0$. Point 2): If $y>0$ then automatically $e^{-lambda y}in(0,1)$. BTW, you haven't answered the questions I posed you in my former comment.
$endgroup$
– drhab
Dec 10 '18 at 18:16
$begingroup$
Thanks again for your answer. I didn't answer your questions because I wanted to make sure I got this straight. So, yes on all questions. Anyway i understood. Using CDF of $X$, which means that $0<e^{-lambda y}leq 1$, i can write $P(Xgeq e^{-lambda y})=int_{e^{-lambda y}}^{infty}f_X(x)dx=int_{e^{-lambda y}}^{1}1dx+int_{1}^{infty}0dx=int_{e^{-lambda y}}^{1}1dx=[x]_{e^{-lambda y}}^{1}=1-e^{-lambda y}$ or $1-P(Xleq e^{-lambda y})=1-int_{-infty}^{e^{-lambda y}}f_X(x)dx=1-[int_{-infty}^{0}0dx+int_{0}^{e^{-lambda y}}1dx]=1-[x]_{0}^{e^{-lambda y}}=1-e^{-lambda y}$.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:39
|
show 3 more comments
$begingroup$
Thank you very much for your answer. I don't understand why, if $F_X(f)$ is defined for $x in [0,1]$, so $F_Y(y)=1-F_X(e^{-lambda y})=1-e^{-lambda y}$. I know that logarithm is defined only for positive values: in fact, for both $x<0$ and $x=0$ we know that $Ynotin mathbb{R}$. Moreover, if $Xin [0,1]$ e $Y$ is a linear transformation of $X$, so $e^{-lambda y}in [0,1]$. But i don't understand how this help me to formalize the passage above.
$endgroup$
– Marco Pittella
Dec 10 '18 at 7:07
$begingroup$
I fail to see your problem. 1) Do you agree that $F_Y(y)=1-F_X(e^{-lambda y})$ for $y>0$? 2) Do you agree that $e^{-lambda y}in(0,1)$ for $y>0$ and $lambda>0$? 3) Do you agree that $F_X(x)=x$ for $xin(0,1)$? If the answer on all questions is "yes" then what withholds you from drawing the conclusion that $F_Y(y)=1-e^{-lambda y}$ for $y>0$? Further it is evident that $F_Y(y)=0$ for $yleq0$ so the complete CDF of $Y$ has been found.
$endgroup$
– drhab
Dec 10 '18 at 9:19
$begingroup$
1) Why $y>0$? 2) Why $e^{-lambda y}in (0,1)$
$endgroup$
– Marco Pittella
Dec 10 '18 at 15:51
$begingroup$
Point 1): $X$ only takes values in $(0,1)$. Then automatically $Y=-frac1{lambda}ln X$ only takes values in $(0,infty)$. That makes it clear immediately that $F_Y(y)=P(Yleq y)=0$ for $yleq0$.Then to accomplish our effort to find $F_Y(y)$ it remains to find $F_Y(y)$ for $y>0$. Point 2): If $y>0$ then automatically $e^{-lambda y}in(0,1)$. BTW, you haven't answered the questions I posed you in my former comment.
$endgroup$
– drhab
Dec 10 '18 at 18:16
$begingroup$
Thanks again for your answer. I didn't answer your questions because I wanted to make sure I got this straight. So, yes on all questions. Anyway i understood. Using CDF of $X$, which means that $0<e^{-lambda y}leq 1$, i can write $P(Xgeq e^{-lambda y})=int_{e^{-lambda y}}^{infty}f_X(x)dx=int_{e^{-lambda y}}^{1}1dx+int_{1}^{infty}0dx=int_{e^{-lambda y}}^{1}1dx=[x]_{e^{-lambda y}}^{1}=1-e^{-lambda y}$ or $1-P(Xleq e^{-lambda y})=1-int_{-infty}^{e^{-lambda y}}f_X(x)dx=1-[int_{-infty}^{0}0dx+int_{0}^{e^{-lambda y}}1dx]=1-[x]_{0}^{e^{-lambda y}}=1-e^{-lambda y}$.
$endgroup$
– Marco Pittella
Dec 11 '18 at 10:39
$begingroup$
Thank you very much for your answer. I don't understand why, if $F_X(f)$ is defined for $x in [0,1]$, so $F_Y(y)=1-F_X(e^{-lambda y})=1-e^{-lambda y}$. I know that logarithm is defined only for positive values: in fact, for both $x<0$ and $x=0$ we know that $Ynotin mathbb{R}$. Moreover, if $Xin [0,1]$ e $Y$ is a linear transformation of $X$, so $e^{-lambda y}in [0,1]$. But i don't understand how this help me to formalize the passage above.
$endgroup$
– Marco Pittella
Dec 10 '18 at 7:07
$begingroup$
Thank you very much for your answer. I don't understand why, if $F_X(f)$ is defined for $x in [0,1]$, so $F_Y(y)=1-F_X(e^{-lambda y})=1-e^{-lambda y}$. I know that logarithm is defined only for positive values: in fact, for both $x<0$ and $x=0$ we know that $Ynotin mathbb{R}$. Moreover, if $Xin [0,1]$ e $Y$ is a linear transformation of $X$, so $e^{-lambda y}in [0,1]$. But i don't understand how this help me to formalize the passage above.
$endgroup$
– Marco Pittella
Dec 10 '18 at 7:07
$begingroup$
I fail to see your problem. 1) Do you agree that $F_Y(y)=1-F_X(e^{-lambda y})$ for $y>0$? 2) Do you agree that $e^{-lambda y}in(0,1)$ for $y>0$ and $lambda>0$? 3) Do you agree that $F_X(x)=x$ for $xin(0,1)$? If the answer on all questions is "yes" then what withholds you from drawing the conclusion that $F_Y(y)=1-e^{-lambda y}$ for $y>0$? Further it is evident that $F_Y(y)=0$ for $yleq0$ so the complete CDF of $Y$ has been found.
$endgroup$
– drhab
Dec 10 '18 at 9:19
$begingroup$
I fail to see your problem. 1) Do you agree that $F_Y(y)=1-F_X(e^{-lambda y})$ for $y>0$? 2) Do you agree that $e^{-lambda y}in(0,1)$ for $y>0$ and $lambda>0$? 3) Do you agree that $F_X(x)=x$ for $xin(0,1)$? If the answer on all questions is "yes" then what withholds you from drawing the conclusion that $F_Y(y)=1-e^{-lambda y}$ for $y>0$? Further it is evident that $F_Y(y)=0$ for $yleq0$ so the complete CDF of $Y$ has been found.
$endgroup$
– drhab
Dec 10 '18 at 9:19
$begingroup$
1) Why $y>0$? 2) Why $e^{-lambda y}in (0,1)$
$endgroup$
– Marco Pittella
Dec 10 '18 at 15:51
$begingroup$
1) Why $y>0$? 2) Why $e^{-lambda y}in (0,1)$
$endgroup$
– Marco Pittella
Dec 10 '18 at 15:51
$begingroup$
Point 1): $X$ only takes values in $(0,1)$. Then automatically $Y=-frac1{lambda}ln X$ only takes values in $(0,infty)$. That makes it clear immediately that $F_Y(y)=P(Yleq y)=0$ for $yleq0$.Then to accomplish our effort to find $F_Y(y)$ it remains to find $F_Y(y)$ for $y>0$. Point 2): If $y>0$ then automatically $e^{-lambda y}in(0,1)$. BTW, you haven't answered the questions I posed you in my former comment.
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– drhab
Dec 10 '18 at 18:16
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Point 1): $X$ only takes values in $(0,1)$. Then automatically $Y=-frac1{lambda}ln X$ only takes values in $(0,infty)$. That makes it clear immediately that $F_Y(y)=P(Yleq y)=0$ for $yleq0$.Then to accomplish our effort to find $F_Y(y)$ it remains to find $F_Y(y)$ for $y>0$. Point 2): If $y>0$ then automatically $e^{-lambda y}in(0,1)$. BTW, you haven't answered the questions I posed you in my former comment.
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– drhab
Dec 10 '18 at 18:16
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Thanks again for your answer. I didn't answer your questions because I wanted to make sure I got this straight. So, yes on all questions. Anyway i understood. Using CDF of $X$, which means that $0<e^{-lambda y}leq 1$, i can write $P(Xgeq e^{-lambda y})=int_{e^{-lambda y}}^{infty}f_X(x)dx=int_{e^{-lambda y}}^{1}1dx+int_{1}^{infty}0dx=int_{e^{-lambda y}}^{1}1dx=[x]_{e^{-lambda y}}^{1}=1-e^{-lambda y}$ or $1-P(Xleq e^{-lambda y})=1-int_{-infty}^{e^{-lambda y}}f_X(x)dx=1-[int_{-infty}^{0}0dx+int_{0}^{e^{-lambda y}}1dx]=1-[x]_{0}^{e^{-lambda y}}=1-e^{-lambda y}$.
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– Marco Pittella
Dec 11 '18 at 10:39
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Thanks again for your answer. I didn't answer your questions because I wanted to make sure I got this straight. So, yes on all questions. Anyway i understood. Using CDF of $X$, which means that $0<e^{-lambda y}leq 1$, i can write $P(Xgeq e^{-lambda y})=int_{e^{-lambda y}}^{infty}f_X(x)dx=int_{e^{-lambda y}}^{1}1dx+int_{1}^{infty}0dx=int_{e^{-lambda y}}^{1}1dx=[x]_{e^{-lambda y}}^{1}=1-e^{-lambda y}$ or $1-P(Xleq e^{-lambda y})=1-int_{-infty}^{e^{-lambda y}}f_X(x)dx=1-[int_{-infty}^{0}0dx+int_{0}^{e^{-lambda y}}1dx]=1-[x]_{0}^{e^{-lambda y}}=1-e^{-lambda y}$.
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– Marco Pittella
Dec 11 '18 at 10:39
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Edit: $g(X)$ is a monotonic function, so I can apply the law of transformation.
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– Marco Pittella
Dec 11 '18 at 10:41