# Convergence (Hall and Jorgenson)

• Nov 28th 2006, 09:24 AM
sh01by
Convergence (Hall and Jorgenson)
Hall and Jorgenson uses the integral to find the present value...

z = [integral from 0 to infinite] exp(-r s) D(s) ds

Find z, so it's a function of t, when...

D(s) = 2(t - s) / t^2 for 0 <= s <= t, D(s) for s > t. (Linear decrease)

It would be appreciated a whole lot, if anyone could take a look at the above...

Good evening,
Simon DK
• Nov 28th 2006, 09:37 AM
ThePerfectHacker
Quote:

Originally Posted by sh01by
Hall and Jorgenson uses the integral to find the present value...

z = [integral from 0 to infinite] exp(-r s) D(s) ds

Find z, so it's a function of t, when...

D(s) = 2(t - s) / t^2 for 0 <= s <= t, D(s) for s > t. (Linear decrease)

Maybe that should be a "t" and not a "r", you probably made a mistake based on the juxtaposition of the buttons.

$z=\int_0^{\infty} e^{-st}D(s)ds$
----
But I am thinking what the function can be. It seems like a Laplace transform but I never seen anything like this before.
• Nov 28th 2006, 09:47 AM
sh01by
Thanks, but...
Hey, thanks for taking time to look at it...

But - I'm quite sure it's supposed to be ' r '...

Notes on the assignment says...

"Hall and Jorgenson uses the integral

z = [integral from 0 to infinite] exp(-r*s)*D(s) ds

to find the discounted present value, to rate r, by a time-dependent flow of depreciations."

I actually have the answer, but it's getting there I'm uncertain of...

z = (2 / (r*t)) * (1 - (1/(r*t))*(1-exp(-r*t)))

Maybe this helps? But - how do I get there?

Any help is still appreciated...
Thanks though...
• Nov 28th 2006, 01:50 PM
topsquark
Quote:

Originally Posted by sh01by
Hall and Jorgenson uses the integral to find the present value...

z = [integral from 0 to infinite] exp(-r s) D(s) ds

Find z, so it's a function of t, when...

D(s) = 2(t - s) / t^2 for 0 <= s <= t, D(s) for s > t. (Linear decrease)

It would be appreciated a whole lot, if anyone could take a look at the above...

Good evening,
Simon DK

Well, it wouldn't completely surprise me that I goofed up somewhere, but I don't like the form of your answer and it doesn't match my work. Even if I made a mistake somewhere I don't see how you could have a factor that includes $e^{-rt}$.

$z = \int_0^{\infty}ds \left ( \frac{2(t-s)}{t^2} \right ) e^{-rs}$

Since the integration is independent of t:
$z = \frac{2}{t} \int_0^{\infty}ds e^{-rs} - \frac{2}{t^2} \int_0^{\infty}ds \, se^{-rs}$

The first integral is trivial:
$\int_0^{\infty}ds e^{-rs} = -\frac{1}{r}e^{-rs}|_0^{\infty} = \frac{1}{r}$

We may do the second integral using integration by parts:
$\int_0^{\infty}ds \, se^{-rs} = \left [ -\frac{se^{-rs}}{r} \right ]_0^{\infty} - \int_0^{\infty}ds \frac{-1}{r}e^{-rs}$

= $\left [ -\frac{se^{-rs}}{r} \right ]_0^{\infty} - \frac{1}{r^2}e^{-rs}|_0^{\infty}$

= $0 + \frac{1}{r^2} = \frac{1}{r^2}$

So we get:
$z = \frac{2}{t} \cdot \frac{1}{r} - \frac{2}{t^2} \cdot \frac{1}{r^2}$

$z = \frac{2}{rt} \left ( 1 - \frac{1}{rt} \right )$

-Dan
• Nov 29th 2006, 03:39 AM
sh01by
Hi (Lecturer has to be questioned...)
Hi,

Thanks for all the help... It has been a great journey, but I think I have to give in... I'll have to talk to my math teacher later today... I have a lecture in 2 hours, so - maybe he can clarify this mystery and ease the pain? :o

I hope I'll come back to this forum with answers...

Simon DK
• Nov 30th 2006, 04:05 AM
sh01by
Lecturer was little help...
Hi there...

I talked with my lecturer and he left me with a little idea of how to proceed...

It's for a math project, which is due till next Friday, so I have a little time to figure it out... I'll be attending another class tomorrow with a different teacher, so I'm hoping to solve the mystery tomorrow...

But - I'll continue with other parts of the project for now...

I'll be back with news. : o )

Simon DK