# Math Help - Limits when evaluating joint pdfs

1. ## Critical regions question

A bus company is about to start a scheduled service between two towns. Before deciding on an appropriate timetable they do ten trial runs to see how long the journey takes. The times, in minutes, are:

89 92 98 34 98 34 98 98 34 34

95% interval for mean journey time?

2. Originally Posted by bluebiro
I have a joint pdf f(s,t) = 6s, 0<t<s<1

I know that you integrate the joint pdf with respect to the other variable when finding marginal densities, but I don't know what limits to use.

So when I integrate 6s w.r.t. t, are the limits 0 and 1? or t and s? or 1 and t? Or s and 0?

Vice versa, when I integrate 6s w.r.t. s, what are the limits? I integrated it with limits 1 and 0 and got 6 as the marginal density, which is surely wrong.

Draw the region defined by 0 < t < s < 1. I put s on the horizontal axis and t on the vertical axis. From what you draw it should be clear that the marginal pdf's are:

$g(s) = \int_{t = 0}^{t = s} f(s, t) \, dt$.

$h(t) = \int_{s = t}^{s = 1} f(s, t) \, ds$.

3. Thank you. I sort of see now. And how about the limits when I find the marginal densities, do they change?

I calculated the marginal density of s given t as 2s/(1-t^2) and t given s as 1/s, how do I know what the range for which this is valid?

4. ## Am I on the right track? (Joint prob density question)

R.v.s s and t have joint p.d.f f(s,t) = 6s, 0<t<s<1. I worked out their marginal densities as
fs(s) = 6s^2
ft(t) = 3-3t^2

and their conditional densities as
f(t given s) = 1/s
f(s given t) = 6s/(3-3t^2)

(a) for what range are the conditional densities valid?
(b) find (t>0.25 given that s = 0.75)

I substituted numbers into the f(t given s) = 1/s equation to get
1/s = 4/3.

Then I integrated this with respect to t, upper limit 1, lower limit 0.25:
[4/3t]

=

4/3 - 1/3 = 1

giving the answer as 1. Is this right?

thanks

5. Originally Posted by bluebiro
R.v.s s and t have joint p.d.f f(s,t) = 6s, 0<t<s<1. I worked out their marginal densities as
fs(s) = 6s^2
ft(t) = 3-3t^2

and their conditional densities as
f(t given s) = 1/s
f(s given t) = 6s/(3-3t^2)

(a) for what range are the conditional densities valid?
(b) find (t>0.25 given that s = 0.75)

I substituted numbers into the f(t given s) = 1/s equation to get
1/s = 4/3.

Then I integrated this with respect to t, upper limit 1, lower limit 0.25:
[4/3t]

=

4/3 - 1/3 = 1

giving the answer as 1. Is this right?

thanks
I've just noticed: The joint pdf you've been given is not valid - the double integral over the region defined by 0 < t < s < 1 is 2 ....

6. Originally Posted by mr fantastic
I've just noticed: The joint pdf you've been given is not valid - the double integral over the region defined by 0 < t < s < 1 is 2 ....
I get 1 (though I could have made a mistake).

integral of 6s w.r.t t (limits is 3s^2)
and the integral of 3s^2 w.r.t s is s^3, which, with limits 1 and 0, gives 1.

Where have I gone wrong?

7. Originally Posted by bluebiro
I get 1 (though I could have made a mistake).

integral of 6s w.r.t t (limits is 3s^2)
and the integral of 3s^2 w.r.t s is s^3, which, with limits 1 and 0, gives 1.

Where have I gone wrong?
$\int_0^s 6s \, dt = \left[6st\right]_0^s = 6s^2$ and $\int_0^1 6s^2 \, ds = \left[ 2s^3 \right]_0^1 = 2$.

$\int_t^1 6s \, ds = \left[3s^2\right]_t^1 = 3 - 3t^2$ and $\int_0^1 3 - 3t^2 \, dt = \left[ 3t - t^3 \right]_0^1 = 2$.

You have not been given a valid joint pdf.

8. Originally Posted by mr fantastic
$\int_0^s 6s \, dt = \left[6st\right]_0^s = 6s^2$ and $\int_0^1 6s^2 \, ds = \left[ 2s^3 \right]_0^1 = 2$.

$\int_t^1 6s \, ds = \left[3s^2\right]_t^1 = 3 - 3t^2$ and $\int_0^1 3 - 3t^2 \, dt = \left[ 3t - t^3 \right]_0^1 = 2$.

You have not been given a valid joint pdf.
Well the joint pdf I was originally given was f(s,t)= lambda.s, where lambda is a constant, so to calculate lambda I did

$1=\lambda\int_0^1\int_0^tsdsdt$
$1=\frac{\lambda}{2}\int_0^1[s^2]_0^tdt$
$1=\frac{\lambda}{2}\int_0^1[t^2]dt$
$1=\frac{\lambda}{6}[t^3]_0^1$
$1=\frac{\lambda}{6}$
$\therefore\lambda=6$

Which is where I got the 6 from.

9. Originally Posted by bluebiro
Well the joint pdf I was originally given was f(s,t)= lambda.s, where lambda is a constant, so to calculate lambda I did

$1=\lambda\int_0^1\int_0^tsdsdt$
$1=\frac{\lambda}{2}\int_0^1[s^2]_0^tdt$
$1=\frac{\lambda}{2}\int_0^1[t^2]dt$
$1=\frac{\lambda}{6}[t^3]_0^1$
$1=\frac{\lambda}{6}$
$\therefore\lambda=6$

Which is where I got the 6 from.
I assume you're referring to this thread: http://www.mathhelpforum.com/math-he...joint-pdf.html

In that thread you define the support as 0 < s < t <1. This is NOT the same as the support defined in this thread, namely 0 < t < s < 1. The joint pdfs in each thread are therefore very different from each other.