Mixing
Casella exercise 4.58: $P(X/Y leq t)$ with $X$ and $Y$ uniform distribution
$begingroup$
Let $X$, $Y$ and $Z$ independent uniform (0,1) random variables:
Find $P(X/Y leq t)$ and $P(XY leq t)$
I started with: $P(X/Y leq t)$ by using Jacobian:
So, $Z= X/Y$, $W = Y$ , my Jacobian gave me: $w$
$F_{Z,W}(z,w) = w$ .
From now on I started to have some problems to define the interval of $F_{Z}(z)$.
Z as I can see goes from zero to infinity right?
The answer in Casella is: $F_{Z} = 1/2 (t)$ , for $t>1$ and $1/2 - (1-t)$, for $t leq 1$
Any help guys?
probability probability-theory probability-distributions
edited Feb 17 at 14:23
Zhanxiong
8,94911032
asked Jan 26 at 20:16
LauraLaura
3518
$endgroup$
add a comment |
$begingroup$
Let $X$, $Y$ and $Z$ independent uniform (0,1) random variables:
Find $P(X/Y leq t)$ and $P(XY leq t)$
I started with: $P(X/Y leq t)$ by using Jacobian:
So, $Z= X/Y$, $W = Y$ , my Jacobian gave me: $w$
$F_{Z,W}(z,w) = w$ .
From now on I started to have some problems to define the interval of $F_{Z}(z)$.
Z as I can see goes from zero to infinity right?
The answer in Casella is: $F_{Z} = 1/2 (t)$ , for $t>1$ and $1/2 - (1-t)$, for $t leq 1$
Any help guys?
probability probability-theory probability-distributions
edited Feb 17 at 14:23
Zhanxiong
8,94911032
asked Jan 26 at 20:16
LauraLaura
3518
$endgroup$
$begingroup$
Have you tried drawing the sample space and sketching the region where $X/Yleq t$?
$endgroup$
– kccu
Jan 26 at 20:21
add a comment |
$begingroup$
Let $X$, $Y$ and $Z$ independent uniform (0,1) random variables:
Find $P(X/Y leq t)$ and $P(XY leq t)$
I started with: $P(X/Y leq t)$ by using Jacobian:
So, $Z= X/Y$, $W = Y$ , my Jacobian gave me: $w$
$F_{Z,W}(z,w) = w$ .
From now on I started to have some problems to define the interval of $F_{Z}(z)$.
Z as I can see goes from zero to infinity right?
The answer in Casella is: $F_{Z} = 1/2 (t)$ , for $t>1$ and $1/2 - (1-t)$, for $t leq 1$
Any help guys?
probability probability-theory probability-distributions
edited Feb 17 at 14:23
Zhanxiong
8,94911032
asked Jan 26 at 20:16
LauraLaura
3518
$endgroup$
Let $X$, $Y$ and $Z$ independent uniform (0,1) random variables:
Find $P(X/Y leq t)$ and $P(XY leq t)$
I started with: $P(X/Y leq t)$ by using Jacobian:
So, $Z= X/Y$, $W = Y$ , my Jacobian gave me: $w$
$F_{Z,W}(z,w) = w$ .
From now on I started to have some problems to define the interval of $F_{Z}(z)$.
Z as I can see goes from zero to infinity right?
The answer in Casella is: $F_{Z} = 1/2 (t)$ , for $t>1$ and $1/2 - (1-t)$, for $t leq 1$
Any help guys?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
edited Feb 17 at 14:23
Zhanxiong
8,94911032
asked Jan 26 at 20:16
LauraLaura
3518
edited Feb 17 at 14:23
Zhanxiong
8,94911032
asked Jan 26 at 20:16
LauraLaura
3518
edited Feb 17 at 14:23
Zhanxiong
8,94911032
edited Feb 17 at 14:23
Zhanxiong
8,94911032
edited Feb 17 at 14:23
Zhanxiong
8,94911032
8,94911032
asked Jan 26 at 20:16
LauraLaura
3518
asked Jan 26 at 20:16
LauraLaura
3518
asked Jan 26 at 20:16
LauraLaura
3518
3518
$begingroup$
Have you tried drawing the sample space and sketching the region where $X/Yleq t$?
$endgroup$
– kccu
Jan 26 at 20:21
add a comment |
$begingroup$
Have you tried drawing the sample space and sketching the region where $X/Yleq t$?
$endgroup$
– kccu
Jan 26 at 20:21
$begingroup$
Have you tried drawing the sample space and sketching the region where $X/Yleq t$?
$endgroup$
– kccu
Jan 26 at 20:21
$begingroup$
Have you tried drawing the sample space and sketching the region where $X/Yleq t$?
$endgroup$
– kccu
Jan 26 at 20:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your way certainly works, but a direct calculation is also possible. First note that the range of $X/Y$ is $mathbb{R}^*$ (the set of all positive reals). The subtleties of this problem is probably to notice the probability depends on the value of $t$. If $t leq 0$, then clearly it is $0$. If $t in (0, 1)$, then $yt in (0, 1)$ for all $y in (0, 1)$, hence
$$P(X/Y leq t) = int_0^1 P(X/y leq t) f_Y(y) dy = int_0^1 P(X leq yt) dy = int_0^1 yt dy = frac{1}{2}t.$$
Similar argument shows that for any $t > 1$,
begin{align}
P(X / Y leq t) & = int_0^1 P(X/y leq t) f_Y(y)dy = int_0^1 P(X leq yt) dy \
& = int_0^{t^{-1}} yt dy + int_{t^{-1}}^1 dy = frac{1}{2}t^{-1} + (1 - t^{-1}) = 1 - frac{1}{2}t^{-1}.
end{align}
Using the same technique (solve the problem by parts), you may try to recover Casella's answer to $F_Z(t)$.
answered Jan 26 at 20:38
ZhanxiongZhanxiong
8,94911032
$endgroup$
$begingroup$
Many many thanks! But this is really confused to me. It is strange to set $Y=y $ and do $X/y leq t$. I will think better about this today.
$endgroup$
– Laura
Jan 26 at 21:11
$begingroup$
In the first Integral you are Integrate the joint distribution of (x,y) right? Something like this: $P(X<yt, 0<Y<1)$ Right? And, of course, $P(X/y <t)$ is a Integral ?
$endgroup$
– Laura
Jan 26 at 21:14
1
$begingroup$
@Laura In general, when two random variables are independent, then for any bivariate function $F$, we have $P(F(X, Y) leq t) = int F(X, y) f_Y(y) dy$ (make analogy of this formula to the (discrete) law of total probability). In this question, simply take $F(x, y) = x/y$.
$endgroup$
– Zhanxiong
Jan 26 at 22:20
1
$begingroup$
Alternatively, as you suggested in your second comment, we can integrate with respect to the joint distribution of $(X, Y)$, which is uniform over $(0, 1) times (0, 1)$. This is slightly more convoluted (in my opinion) -- you need to sketch the region ${(x, y): x/y leq t}$ then do the double integration over $(0, 1) times (0, 1)$. This region, of course, also depends on $t$. Note this method is somewhat different from my answer. You can try it out by yourself --- if you encountered any difficulties, I can update my current answer.
$endgroup$
– Zhanxiong
Jan 26 at 22:27
1
$begingroup$
@Zhanxiong Yes yes! I got it now! The independence hypothesis allows me to do that! Thanks
$endgroup$
– Laura
Jan 26 at 23:17
|
show 1 more comment
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1 Answer
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1 Answer
1
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oldest
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oldest
votes
$begingroup$
Your way certainly works, but a direct calculation is also possible. First note that the range of $X/Y$ is $mathbb{R}^*$ (the set of all positive reals). The subtleties of this problem is probably to notice the probability depends on the value of $t$. If $t leq 0$, then clearly it is $0$. If $t in (0, 1)$, then $yt in (0, 1)$ for all $y in (0, 1)$, hence
$$P(X/Y leq t) = int_0^1 P(X/y leq t) f_Y(y) dy = int_0^1 P(X leq yt) dy = int_0^1 yt dy = frac{1}{2}t.$$
Similar argument shows that for any $t > 1$,
begin{align}
P(X / Y leq t) & = int_0^1 P(X/y leq t) f_Y(y)dy = int_0^1 P(X leq yt) dy \
& = int_0^{t^{-1}} yt dy + int_{t^{-1}}^1 dy = frac{1}{2}t^{-1} + (1 - t^{-1}) = 1 - frac{1}{2}t^{-1}.
end{align}
Using the same technique (solve the problem by parts), you may try to recover Casella's answer to $F_Z(t)$.
answered Jan 26 at 20:38
ZhanxiongZhanxiong
8,94911032
$endgroup$
$begingroup$
Many many thanks! But this is really confused to me. It is strange to set $Y=y $ and do $X/y leq t$. I will think better about this today.
$endgroup$
– Laura
Jan 26 at 21:11
$begingroup$
In the first Integral you are Integrate the joint distribution of (x,y) right? Something like this: $P(X<yt, 0<Y<1)$ Right? And, of course, $P(X/y <t)$ is a Integral ?
$endgroup$
– Laura
Jan 26 at 21:14
1
$begingroup$
@Laura In general, when two random variables are independent, then for any bivariate function $F$, we have $P(F(X, Y) leq t) = int F(X, y) f_Y(y) dy$ (make analogy of this formula to the (discrete) law of total probability). In this question, simply take $F(x, y) = x/y$.
$endgroup$
– Zhanxiong
Jan 26 at 22:20
1
$begingroup$
Alternatively, as you suggested in your second comment, we can integrate with respect to the joint distribution of $(X, Y)$, which is uniform over $(0, 1) times (0, 1)$. This is slightly more convoluted (in my opinion) -- you need to sketch the region ${(x, y): x/y leq t}$ then do the double integration over $(0, 1) times (0, 1)$. This region, of course, also depends on $t$. Note this method is somewhat different from my answer. You can try it out by yourself --- if you encountered any difficulties, I can update my current answer.
$endgroup$
– Zhanxiong
Jan 26 at 22:27
1
$begingroup$
@Zhanxiong Yes yes! I got it now! The independence hypothesis allows me to do that! Thanks
$endgroup$
– Laura
Jan 26 at 23:17
|
show 1 more comment
$begingroup$
Your way certainly works, but a direct calculation is also possible. First note that the range of $X/Y$ is $mathbb{R}^*$ (the set of all positive reals). The subtleties of this problem is probably to notice the probability depends on the value of $t$. If $t leq 0$, then clearly it is $0$. If $t in (0, 1)$, then $yt in (0, 1)$ for all $y in (0, 1)$, hence
$$P(X/Y leq t) = int_0^1 P(X/y leq t) f_Y(y) dy = int_0^1 P(X leq yt) dy = int_0^1 yt dy = frac{1}{2}t.$$
Similar argument shows that for any $t > 1$,
begin{align}
P(X / Y leq t) & = int_0^1 P(X/y leq t) f_Y(y)dy = int_0^1 P(X leq yt) dy \
& = int_0^{t^{-1}} yt dy + int_{t^{-1}}^1 dy = frac{1}{2}t^{-1} + (1 - t^{-1}) = 1 - frac{1}{2}t^{-1}.
end{align}
Using the same technique (solve the problem by parts), you may try to recover Casella's answer to $F_Z(t)$.
answered Jan 26 at 20:38
ZhanxiongZhanxiong
8,94911032
$endgroup$
$begingroup$
Many many thanks! But this is really confused to me. It is strange to set $Y=y $ and do $X/y leq t$. I will think better about this today.
$endgroup$
– Laura
Jan 26 at 21:11
$begingroup$
In the first Integral you are Integrate the joint distribution of (x,y) right? Something like this: $P(X<yt, 0<Y<1)$ Right? And, of course, $P(X/y <t)$ is a Integral ?
$endgroup$
– Laura
Jan 26 at 21:14
1
$begingroup$
@Laura In general, when two random variables are independent, then for any bivariate function $F$, we have $P(F(X, Y) leq t) = int F(X, y) f_Y(y) dy$ (make analogy of this formula to the (discrete) law of total probability). In this question, simply take $F(x, y) = x/y$.
$endgroup$
– Zhanxiong
Jan 26 at 22:20
1
$begingroup$
Alternatively, as you suggested in your second comment, we can integrate with respect to the joint distribution of $(X, Y)$, which is uniform over $(0, 1) times (0, 1)$. This is slightly more convoluted (in my opinion) -- you need to sketch the region ${(x, y): x/y leq t}$ then do the double integration over $(0, 1) times (0, 1)$. This region, of course, also depends on $t$. Note this method is somewhat different from my answer. You can try it out by yourself --- if you encountered any difficulties, I can update my current answer.
$endgroup$
– Zhanxiong
Jan 26 at 22:27
1
$begingroup$
@Zhanxiong Yes yes! I got it now! The independence hypothesis allows me to do that! Thanks
$endgroup$
– Laura
Jan 26 at 23:17
|
show 1 more comment
$begingroup$
Your way certainly works, but a direct calculation is also possible. First note that the range of $X/Y$ is $mathbb{R}^*$ (the set of all positive reals). The subtleties of this problem is probably to notice the probability depends on the value of $t$. If $t leq 0$, then clearly it is $0$. If $t in (0, 1)$, then $yt in (0, 1)$ for all $y in (0, 1)$, hence
$$P(X/Y leq t) = int_0^1 P(X/y leq t) f_Y(y) dy = int_0^1 P(X leq yt) dy = int_0^1 yt dy = frac{1}{2}t.$$
Similar argument shows that for any $t > 1$,
begin{align}
P(X / Y leq t) & = int_0^1 P(X/y leq t) f_Y(y)dy = int_0^1 P(X leq yt) dy \
& = int_0^{t^{-1}} yt dy + int_{t^{-1}}^1 dy = frac{1}{2}t^{-1} + (1 - t^{-1}) = 1 - frac{1}{2}t^{-1}.
end{align}
Using the same technique (solve the problem by parts), you may try to recover Casella's answer to $F_Z(t)$.
answered Jan 26 at 20:38
ZhanxiongZhanxiong
8,94911032
$endgroup$
Your way certainly works, but a direct calculation is also possible. First note that the range of $X/Y$ is $mathbb{R}^*$ (the set of all positive reals). The subtleties of this problem is probably to notice the probability depends on the value of $t$. If $t leq 0$, then clearly it is $0$. If $t in (0, 1)$, then $yt in (0, 1)$ for all $y in (0, 1)$, hence
$$P(X/Y leq t) = int_0^1 P(X/y leq t) f_Y(y) dy = int_0^1 P(X leq yt) dy = int_0^1 yt dy = frac{1}{2}t.$$
Similar argument shows that for any $t > 1$,
begin{align}
P(X / Y leq t) & = int_0^1 P(X/y leq t) f_Y(y)dy = int_0^1 P(X leq yt) dy \
& = int_0^{t^{-1}} yt dy + int_{t^{-1}}^1 dy = frac{1}{2}t^{-1} + (1 - t^{-1}) = 1 - frac{1}{2}t^{-1}.
end{align}
Using the same technique (solve the problem by parts), you may try to recover Casella's answer to $F_Z(t)$.
answered Jan 26 at 20:38
ZhanxiongZhanxiong
8,94911032
answered Jan 26 at 20:38
ZhanxiongZhanxiong
8,94911032
answered Jan 26 at 20:38
ZhanxiongZhanxiong
8,94911032
answered Jan 26 at 20:38
ZhanxiongZhanxiong
8,94911032
8,94911032
$begingroup$
Many many thanks! But this is really confused to me. It is strange to set $Y=y $ and do $X/y leq t$. I will think better about this today.
$endgroup$
– Laura
Jan 26 at 21:11
$begingroup$
In the first Integral you are Integrate the joint distribution of (x,y) right? Something like this: $P(X<yt, 0<Y<1)$ Right? And, of course, $P(X/y <t)$ is a Integral ?
$endgroup$
– Laura
Jan 26 at 21:14
1
$begingroup$
@Laura In general, when two random variables are independent, then for any bivariate function $F$, we have $P(F(X, Y) leq t) = int F(X, y) f_Y(y) dy$ (make analogy of this formula to the (discrete) law of total probability). In this question, simply take $F(x, y) = x/y$.
$endgroup$
– Zhanxiong
Jan 26 at 22:20
1
$begingroup$
Alternatively, as you suggested in your second comment, we can integrate with respect to the joint distribution of $(X, Y)$, which is uniform over $(0, 1) times (0, 1)$. This is slightly more convoluted (in my opinion) -- you need to sketch the region ${(x, y): x/y leq t}$ then do the double integration over $(0, 1) times (0, 1)$. This region, of course, also depends on $t$. Note this method is somewhat different from my answer. You can try it out by yourself --- if you encountered any difficulties, I can update my current answer.
$endgroup$
– Zhanxiong
Jan 26 at 22:27
1
$begingroup$
@Zhanxiong Yes yes! I got it now! The independence hypothesis allows me to do that! Thanks
$endgroup$
– Laura
Jan 26 at 23:17
|
show 1 more comment
$begingroup$
Many many thanks! But this is really confused to me. It is strange to set $Y=y $ and do $X/y leq t$. I will think better about this today.
$endgroup$
– Laura
Jan 26 at 21:11
$begingroup$
In the first Integral you are Integrate the joint distribution of (x,y) right? Something like this: $P(X<yt, 0<Y<1)$ Right? And, of course, $P(X/y <t)$ is a Integral ?
$endgroup$
– Laura
Jan 26 at 21:14
1
$begingroup$
@Laura In general, when two random variables are independent, then for any bivariate function $F$, we have $P(F(X, Y) leq t) = int F(X, y) f_Y(y) dy$ (make analogy of this formula to the (discrete) law of total probability). In this question, simply take $F(x, y) = x/y$.
$endgroup$
– Zhanxiong
Jan 26 at 22:20
1
$begingroup$
Alternatively, as you suggested in your second comment, we can integrate with respect to the joint distribution of $(X, Y)$, which is uniform over $(0, 1) times (0, 1)$. This is slightly more convoluted (in my opinion) -- you need to sketch the region ${(x, y): x/y leq t}$ then do the double integration over $(0, 1) times (0, 1)$. This region, of course, also depends on $t$. Note this method is somewhat different from my answer. You can try it out by yourself --- if you encountered any difficulties, I can update my current answer.
$endgroup$
– Zhanxiong
Jan 26 at 22:27
1
$begingroup$
@Zhanxiong Yes yes! I got it now! The independence hypothesis allows me to do that! Thanks
$endgroup$
– Laura
Jan 26 at 23:17
$begingroup$
Many many thanks! But this is really confused to me. It is strange to set $Y=y $ and do $X/y leq t$. I will think better about this today.
$endgroup$
– Laura
Jan 26 at 21:11
$begingroup$
Many many thanks! But this is really confused to me. It is strange to set $Y=y $ and do $X/y leq t$. I will think better about this today.
$endgroup$
– Laura
Jan 26 at 21:11
$begingroup$
In the first Integral you are Integrate the joint distribution of (x,y) right? Something like this: $P(X<yt, 0<Y<1)$ Right? And, of course, $P(X/y <t)$ is a Integral ?
$endgroup$
– Laura
Jan 26 at 21:14
$begingroup$
In the first Integral you are Integrate the joint distribution of (x,y) right? Something like this: $P(X<yt, 0<Y<1)$ Right? And, of course, $P(X/y <t)$ is a Integral ?
$endgroup$
– Laura
Jan 26 at 21:14
1
1
$begingroup$
@Laura In general, when two random variables are independent, then for any bivariate function $F$, we have $P(F(X, Y) leq t) = int F(X, y) f_Y(y) dy$ (make analogy of this formula to the (discrete) law of total probability). In this question, simply take $F(x, y) = x/y$.
$endgroup$
– Zhanxiong
Jan 26 at 22:20
$begingroup$
@Laura In general, when two random variables are independent, then for any bivariate function $F$, we have $P(F(X, Y) leq t) = int F(X, y) f_Y(y) dy$ (make analogy of this formula to the (discrete) law of total probability). In this question, simply take $F(x, y) = x/y$.
$endgroup$
– Zhanxiong
Jan 26 at 22:20
1
1
$begingroup$
Alternatively, as you suggested in your second comment, we can integrate with respect to the joint distribution of $(X, Y)$, which is uniform over $(0, 1) times (0, 1)$. This is slightly more convoluted (in my opinion) -- you need to sketch the region ${(x, y): x/y leq t}$ then do the double integration over $(0, 1) times (0, 1)$. This region, of course, also depends on $t$. Note this method is somewhat different from my answer. You can try it out by yourself --- if you encountered any difficulties, I can update my current answer.
$endgroup$
– Zhanxiong
Jan 26 at 22:27
$begingroup$
Alternatively, as you suggested in your second comment, we can integrate with respect to the joint distribution of $(X, Y)$, which is uniform over $(0, 1) times (0, 1)$. This is slightly more convoluted (in my opinion) -- you need to sketch the region ${(x, y): x/y leq t}$ then do the double integration over $(0, 1) times (0, 1)$. This region, of course, also depends on $t$. Note this method is somewhat different from my answer. You can try it out by yourself --- if you encountered any difficulties, I can update my current answer.
$endgroup$
– Zhanxiong
Jan 26 at 22:27
1
1
$begingroup$
@Zhanxiong Yes yes! I got it now! The independence hypothesis allows me to do that! Thanks
$endgroup$
– Laura
Jan 26 at 23:17
$begingroup$
@Zhanxiong Yes yes! I got it now! The independence hypothesis allows me to do that! Thanks
$endgroup$
– Laura
Jan 26 at 23:17
|
show 1 more comment
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$begingroup$
Have you tried drawing the sample space and sketching the region where $X/Yleq t$?
$endgroup$
– kccu
Jan 26 at 20:21