Mixing
Given $ X sim U[-1, 1]$ find the PDF of $Y=frac{1}{1+X}$ - how to divide to ranges?
$begingroup$
Regarding the following problem:
Given $ X sim U[-1, 1]$ find the PDF of $Y=frac{1}{1+X}$
So:
$F_Y(t)=P(Yleq t)=P(frac{1}{1+X}leq t)=P(X> frac{1-t}{t}) = 1 - P(Xleq frac{1-t}{t})$
Now I need to seprate the different cases for t - and this is the point I always fail to do. I tried to plot the function:
And to extract information from there, and the only ranges I could think of were $t<0$ and $tgeq 0$ which is probably incorrect.
Could someone please explain me how to approach such problems?
Thanks
probability probability-distributions
asked Jan 30 at 12:46
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
Regarding the following problem:
Given $ X sim U[-1, 1]$ find the PDF of $Y=frac{1}{1+X}$
So:
$F_Y(t)=P(Yleq t)=P(frac{1}{1+X}leq t)=P(X> frac{1-t}{t}) = 1 - P(Xleq frac{1-t}{t})$
Now I need to seprate the different cases for t - and this is the point I always fail to do. I tried to plot the function:
And to extract information from there, and the only ranges I could think of were $t<0$ and $tgeq 0$ which is probably incorrect.
Could someone please explain me how to approach such problems?
Thanks
probability probability-distributions
asked Jan 30 at 12:46
superuser123superuser123
48628
$endgroup$
$begingroup$
Why not just use a change of variables ? $Y=frac{1}{1+X}implies x=frac{1-y}{y}$ and $-1<x<1implies 0<1+x<2implies y>frac{1}{2}$. So the pdf of $Y$ is $$f_Y(y)=f_Xleft(frac{1-y}{y}right)left|frac{dx}{dy}right|mathbf1_{y>1/2}=cdots$$
$endgroup$
– StubbornAtom
Jan 30 at 15:01
add a comment |
$begingroup$
Regarding the following problem:
Given $ X sim U[-1, 1]$ find the PDF of $Y=frac{1}{1+X}$
So:
$F_Y(t)=P(Yleq t)=P(frac{1}{1+X}leq t)=P(X> frac{1-t}{t}) = 1 - P(Xleq frac{1-t}{t})$
Now I need to seprate the different cases for t - and this is the point I always fail to do. I tried to plot the function:
And to extract information from there, and the only ranges I could think of were $t<0$ and $tgeq 0$ which is probably incorrect.
Could someone please explain me how to approach such problems?
Thanks
probability probability-distributions
asked Jan 30 at 12:46
superuser123superuser123
48628
$endgroup$
Regarding the following problem:
Given $ X sim U[-1, 1]$ find the PDF of $Y=frac{1}{1+X}$
So:
$F_Y(t)=P(Yleq t)=P(frac{1}{1+X}leq t)=P(X> frac{1-t}{t}) = 1 - P(Xleq frac{1-t}{t})$
Now I need to seprate the different cases for t - and this is the point I always fail to do. I tried to plot the function:
And to extract information from there, and the only ranges I could think of were $t<0$ and $tgeq 0$ which is probably incorrect.
Could someone please explain me how to approach such problems?
Thanks
probability probability-distributions
probability probability-distributions
asked Jan 30 at 12:46
superuser123superuser123
48628
asked Jan 30 at 12:46
superuser123superuser123
48628
asked Jan 30 at 12:46
superuser123superuser123
48628
asked Jan 30 at 12:46
superuser123superuser123
48628
asked Jan 30 at 12:46
superuser123superuser123
48628
48628
$begingroup$
Why not just use a change of variables ? $Y=frac{1}{1+X}implies x=frac{1-y}{y}$ and $-1<x<1implies 0<1+x<2implies y>frac{1}{2}$. So the pdf of $Y$ is $$f_Y(y)=f_Xleft(frac{1-y}{y}right)left|frac{dx}{dy}right|mathbf1_{y>1/2}=cdots$$
$endgroup$
– StubbornAtom
Jan 30 at 15:01
add a comment |
$begingroup$
Why not just use a change of variables ? $Y=frac{1}{1+X}implies x=frac{1-y}{y}$ and $-1<x<1implies 0<1+x<2implies y>frac{1}{2}$. So the pdf of $Y$ is $$f_Y(y)=f_Xleft(frac{1-y}{y}right)left|frac{dx}{dy}right|mathbf1_{y>1/2}=cdots$$
$endgroup$
– StubbornAtom
Jan 30 at 15:01
$begingroup$
Why not just use a change of variables ? $Y=frac{1}{1+X}implies x=frac{1-y}{y}$ and $-1<x<1implies 0<1+x<2implies y>frac{1}{2}$. So the pdf of $Y$ is $$f_Y(y)=f_Xleft(frac{1-y}{y}right)left|frac{dx}{dy}right|mathbf1_{y>1/2}=cdots$$
$endgroup$
– StubbornAtom
Jan 30 at 15:01
$begingroup$
Why not just use a change of variables ? $Y=frac{1}{1+X}implies x=frac{1-y}{y}$ and $-1<x<1implies 0<1+x<2implies y>frac{1}{2}$. So the pdf of $Y$ is $$f_Y(y)=f_Xleft(frac{1-y}{y}right)left|frac{dx}{dy}right|mathbf1_{y>1/2}=cdots$$
$endgroup$
– StubbornAtom
Jan 30 at 15:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Observe that $1+X in [0,2]$ so that $Y in [1/2,+infty)$ almost surely. This means that for $t leq 1/2$ you have $F_Y(t) = 0$. The expression you wrote holds for $t>1/2$.
In general, your procedure is correct. You just have to take care of where your random variable takes values. If, for example, $Z in [a,b]$ then you'll have $F_Z(x) = 0$ for $x < a$ and $F_Z(x) = 1$ for $x geq b$.
Also, remind to take care of what happens when you divide. In this case, there's a point where you divide by $t$ but that would change the sign of the inequality. However, we are sure that $t geq 1/2$ for the previous discussion, so there's no need to make this distinction now.
Side note: how did I calculate the support of $Y$? I calculated the image of
$$f(x) = frac{1}{1+x}, quad x in
[-1,1]$$
Since $f$ is decreasing and continuous, it suffices to observe that $$f(1)=1/2$$
$$lim_{x to -1^+} f(x) = + infty$$
to get the desired image.
answered Jan 30 at 12:56
HarnakHarnak
1,344512
$endgroup$
$begingroup$
So these are the only two ranges? because when $tgeqfrac{1}{2}$ it seems like it behaves different then $0<t<frac{1}{2}$
$endgroup$
– superuser123
Jan 30 at 12:58
$begingroup$
How does it behave differently? What happens for those ranges?
$endgroup$
– Harnak
Jan 30 at 13:03
$begingroup$
If $tgeq frac{1}{2}$ then $-1 leq frac{1-t}{t} leq 2 $ , which is in the range where $F_X(frac{1-t}{t}) = frac{t+1}{3}$
$endgroup$
– superuser123
Jan 30 at 13:06
$begingroup$
Sorry, actually, I made a slight mistake in the calculation. I'll edit this.
$endgroup$
– Harnak
Jan 30 at 13:09
1
$begingroup$
@superuser123, I'm a bit busy at work right now, but I'll add those steps later :)
$endgroup$
– Harnak
Jan 30 at 13:41
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Observe that $1+X in [0,2]$ so that $Y in [1/2,+infty)$ almost surely. This means that for $t leq 1/2$ you have $F_Y(t) = 0$. The expression you wrote holds for $t>1/2$.
In general, your procedure is correct. You just have to take care of where your random variable takes values. If, for example, $Z in [a,b]$ then you'll have $F_Z(x) = 0$ for $x < a$ and $F_Z(x) = 1$ for $x geq b$.
Also, remind to take care of what happens when you divide. In this case, there's a point where you divide by $t$ but that would change the sign of the inequality. However, we are sure that $t geq 1/2$ for the previous discussion, so there's no need to make this distinction now.
Side note: how did I calculate the support of $Y$? I calculated the image of
$$f(x) = frac{1}{1+x}, quad x in
[-1,1]$$
Since $f$ is decreasing and continuous, it suffices to observe that $$f(1)=1/2$$
$$lim_{x to -1^+} f(x) = + infty$$
to get the desired image.
answered Jan 30 at 12:56
HarnakHarnak
1,344512
$endgroup$
$begingroup$
So these are the only two ranges? because when $tgeqfrac{1}{2}$ it seems like it behaves different then $0<t<frac{1}{2}$
$endgroup$
– superuser123
Jan 30 at 12:58
$begingroup$
How does it behave differently? What happens for those ranges?
$endgroup$
– Harnak
Jan 30 at 13:03
$begingroup$
If $tgeq frac{1}{2}$ then $-1 leq frac{1-t}{t} leq 2 $ , which is in the range where $F_X(frac{1-t}{t}) = frac{t+1}{3}$
$endgroup$
– superuser123
Jan 30 at 13:06
$begingroup$
Sorry, actually, I made a slight mistake in the calculation. I'll edit this.
$endgroup$
– Harnak
Jan 30 at 13:09
1
$begingroup$
@superuser123, I'm a bit busy at work right now, but I'll add those steps later :)
$endgroup$
– Harnak
Jan 30 at 13:41
|
show 3 more comments
$begingroup$
Observe that $1+X in [0,2]$ so that $Y in [1/2,+infty)$ almost surely. This means that for $t leq 1/2$ you have $F_Y(t) = 0$. The expression you wrote holds for $t>1/2$.
In general, your procedure is correct. You just have to take care of where your random variable takes values. If, for example, $Z in [a,b]$ then you'll have $F_Z(x) = 0$ for $x < a$ and $F_Z(x) = 1$ for $x geq b$.
Also, remind to take care of what happens when you divide. In this case, there's a point where you divide by $t$ but that would change the sign of the inequality. However, we are sure that $t geq 1/2$ for the previous discussion, so there's no need to make this distinction now.
Side note: how did I calculate the support of $Y$? I calculated the image of
$$f(x) = frac{1}{1+x}, quad x in
[-1,1]$$
Since $f$ is decreasing and continuous, it suffices to observe that $$f(1)=1/2$$
$$lim_{x to -1^+} f(x) = + infty$$
to get the desired image.
answered Jan 30 at 12:56
HarnakHarnak
1,344512
$endgroup$
$begingroup$
So these are the only two ranges? because when $tgeqfrac{1}{2}$ it seems like it behaves different then $0<t<frac{1}{2}$
$endgroup$
– superuser123
Jan 30 at 12:58
$begingroup$
How does it behave differently? What happens for those ranges?
$endgroup$
– Harnak
Jan 30 at 13:03
$begingroup$
If $tgeq frac{1}{2}$ then $-1 leq frac{1-t}{t} leq 2 $ , which is in the range where $F_X(frac{1-t}{t}) = frac{t+1}{3}$
$endgroup$
– superuser123
Jan 30 at 13:06
$begingroup$
Sorry, actually, I made a slight mistake in the calculation. I'll edit this.
$endgroup$
– Harnak
Jan 30 at 13:09
1
$begingroup$
@superuser123, I'm a bit busy at work right now, but I'll add those steps later :)
$endgroup$
– Harnak
Jan 30 at 13:41
|
show 3 more comments
$begingroup$
Observe that $1+X in [0,2]$ so that $Y in [1/2,+infty)$ almost surely. This means that for $t leq 1/2$ you have $F_Y(t) = 0$. The expression you wrote holds for $t>1/2$.
In general, your procedure is correct. You just have to take care of where your random variable takes values. If, for example, $Z in [a,b]$ then you'll have $F_Z(x) = 0$ for $x < a$ and $F_Z(x) = 1$ for $x geq b$.
Also, remind to take care of what happens when you divide. In this case, there's a point where you divide by $t$ but that would change the sign of the inequality. However, we are sure that $t geq 1/2$ for the previous discussion, so there's no need to make this distinction now.
Side note: how did I calculate the support of $Y$? I calculated the image of
$$f(x) = frac{1}{1+x}, quad x in
[-1,1]$$
Since $f$ is decreasing and continuous, it suffices to observe that $$f(1)=1/2$$
$$lim_{x to -1^+} f(x) = + infty$$
to get the desired image.
answered Jan 30 at 12:56
HarnakHarnak
1,344512
$endgroup$
Observe that $1+X in [0,2]$ so that $Y in [1/2,+infty)$ almost surely. This means that for $t leq 1/2$ you have $F_Y(t) = 0$. The expression you wrote holds for $t>1/2$.
In general, your procedure is correct. You just have to take care of where your random variable takes values. If, for example, $Z in [a,b]$ then you'll have $F_Z(x) = 0$ for $x < a$ and $F_Z(x) = 1$ for $x geq b$.
Also, remind to take care of what happens when you divide. In this case, there's a point where you divide by $t$ but that would change the sign of the inequality. However, we are sure that $t geq 1/2$ for the previous discussion, so there's no need to make this distinction now.
Side note: how did I calculate the support of $Y$? I calculated the image of
$$f(x) = frac{1}{1+x}, quad x in
[-1,1]$$
Since $f$ is decreasing and continuous, it suffices to observe that $$f(1)=1/2$$
$$lim_{x to -1^+} f(x) = + infty$$
to get the desired image.
answered Jan 30 at 12:56
HarnakHarnak
1,344512
edited Jan 30 at 15:17
answered Jan 30 at 12:56
HarnakHarnak
1,344512
answered Jan 30 at 12:56
HarnakHarnak
1,344512
answered Jan 30 at 12:56
HarnakHarnak
1,344512
1,344512
$begingroup$
So these are the only two ranges? because when $tgeqfrac{1}{2}$ it seems like it behaves different then $0<t<frac{1}{2}$
$endgroup$
– superuser123
Jan 30 at 12:58
$begingroup$
How does it behave differently? What happens for those ranges?
$endgroup$
– Harnak
Jan 30 at 13:03
$begingroup$
If $tgeq frac{1}{2}$ then $-1 leq frac{1-t}{t} leq 2 $ , which is in the range where $F_X(frac{1-t}{t}) = frac{t+1}{3}$
$endgroup$
– superuser123
Jan 30 at 13:06
$begingroup$
Sorry, actually, I made a slight mistake in the calculation. I'll edit this.
$endgroup$
– Harnak
Jan 30 at 13:09
1
$begingroup$
@superuser123, I'm a bit busy at work right now, but I'll add those steps later :)
$endgroup$
– Harnak
Jan 30 at 13:41
|
show 3 more comments
$begingroup$
So these are the only two ranges? because when $tgeqfrac{1}{2}$ it seems like it behaves different then $0<t<frac{1}{2}$
$endgroup$
– superuser123
Jan 30 at 12:58
$begingroup$
How does it behave differently? What happens for those ranges?
$endgroup$
– Harnak
Jan 30 at 13:03
$begingroup$
If $tgeq frac{1}{2}$ then $-1 leq frac{1-t}{t} leq 2 $ , which is in the range where $F_X(frac{1-t}{t}) = frac{t+1}{3}$
$endgroup$
– superuser123
Jan 30 at 13:06
$begingroup$
Sorry, actually, I made a slight mistake in the calculation. I'll edit this.
$endgroup$
– Harnak
Jan 30 at 13:09
1
$begingroup$
@superuser123, I'm a bit busy at work right now, but I'll add those steps later :)
$endgroup$
– Harnak
Jan 30 at 13:41
$begingroup$
So these are the only two ranges? because when $tgeqfrac{1}{2}$ it seems like it behaves different then $0<t<frac{1}{2}$
$endgroup$
– superuser123
Jan 30 at 12:58
$begingroup$
So these are the only two ranges? because when $tgeqfrac{1}{2}$ it seems like it behaves different then $0<t<frac{1}{2}$
$endgroup$
– superuser123
Jan 30 at 12:58
$begingroup$
How does it behave differently? What happens for those ranges?
$endgroup$
– Harnak
Jan 30 at 13:03
$begingroup$
How does it behave differently? What happens for those ranges?
$endgroup$
– Harnak
Jan 30 at 13:03
$begingroup$
If $tgeq frac{1}{2}$ then $-1 leq frac{1-t}{t} leq 2 $ , which is in the range where $F_X(frac{1-t}{t}) = frac{t+1}{3}$
$endgroup$
– superuser123
Jan 30 at 13:06
$begingroup$
If $tgeq frac{1}{2}$ then $-1 leq frac{1-t}{t} leq 2 $ , which is in the range where $F_X(frac{1-t}{t}) = frac{t+1}{3}$
$endgroup$
– superuser123
Jan 30 at 13:06
$begingroup$
Sorry, actually, I made a slight mistake in the calculation. I'll edit this.
$endgroup$
– Harnak
Jan 30 at 13:09
$begingroup$
Sorry, actually, I made a slight mistake in the calculation. I'll edit this.
$endgroup$
– Harnak
Jan 30 at 13:09
1
1
$begingroup$
@superuser123, I'm a bit busy at work right now, but I'll add those steps later :)
$endgroup$
– Harnak
Jan 30 at 13:41
$begingroup$
@superuser123, I'm a bit busy at work right now, but I'll add those steps later :)
$endgroup$
– Harnak
Jan 30 at 13:41
|
show 3 more comments
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$begingroup$
Why not just use a change of variables ? $Y=frac{1}{1+X}implies x=frac{1-y}{y}$ and $-1<x<1implies 0<1+x<2implies y>frac{1}{2}$. So the pdf of $Y$ is $$f_Y(y)=f_Xleft(frac{1-y}{y}right)left|frac{dx}{dy}right|mathbf1_{y>1/2}=cdots$$
$endgroup$
– StubbornAtom
Jan 30 at 15:01