Is the calculation related to average or probability?

[zak100]

TL;DR Summary

I am trying to understand a calculation in a research. I feel they are trying to calculate probability

They are doing following calculation:
They are doing following calculation:

1/6*(uh1) + 3/6*(uh2) + 2/6 (uh3) =17/6

Hi,

I am trying to understand the calculation in the following research paper:
http://cramton.umd.edu/market-design/abdulkadiroglu-sonmez-house-allocation.pdf

 

[Stephen Tashi]

They are doing following calculation:

1/6*(uh1) + 3/6*(uh2) + 2/6 (uh3) =17/6

According to the paper, the table of utilities gives $u(h_1)= 3$, $u(h_2) =4$, $u(h_3) = 1$. The calculation shown in the paper is $(1/6) u(h_1) +( 3/6)u(h_2) + (2/6) (u(h_3) = 3/6 + 12/6 + 2/6 = 17/6$.

For tenant 1, the function $u$ is a random variable. It has 3 possible outcomes (the $u(h_i)$) and with each outcome there is an associated probability $p_i$ The sum of products of the form $p_i u(h_i)$ is the expected value of the random variable, which some call its average.

 

[zak100]

Hi,
Thanks for your response. I am still reading.
Zulfi.

 

[zak100]

Hi,

Thanks for your response:

with each outcome there is an associated probability $p_i$

Please tell me how we can state this probability $p_i$ in words?

Zulfi.

 

[Stephen Tashi]

Each of the 6 orderings is assumed to have an equal probability. For example, the ordering $i_2, i_1, i_3$ has probability 1/6. With that ordering, we assume tenant $i_2$ picks his favorite house, so he picks house $h_1$. The leaves the choices available for $i_1$ to be houses $h_2,\ h_3$. Tenant $i_1$ choses house $h_2$ since it has utility 4 to him and 4 > 1.

There are 3 out of 6 orderings where tenant $i_1$ choses house $h_2$ and obtains a utility of 4. So there is a 3/6 chance that the random variable $u$ = "utility of the result to tenant 1" is equal to 4.

$p_i$ = the probability that tenant $i_1$ obtains utility $u(h_i)$

 

[zak100]

Thanks a lot for solving my problem.

God bless you.

 

[zak100]

Hi,

Sorry I can't understand the fractional values 1/6, 2/6 and 3/6. How can you say that :

For example, the ordering i2,i1,i3 has probability 1/6.

Somebody please guide me.

Zulfi.

 

[Stephen Tashi]

Sorry I can't understand the fractional values 1/6, 2/6 and 3/6. How can you say that :

The paper assumes all possible orderings of how the 3 tenants make choices have the same probability. There are 6 possible orders: (1,2,3),(1,3,2),(2,1,3),... etc.

Some possible orders result in the same utility for observer 1.