Now Due Wednesday February 17
It has been brought to my attention that even the scaled back version of the problem is way too hard… and I agree and my sincere apologies.
Byron arrives precisely at time 0 and Ludd arrives with a uniform probability distribution from -1 to 1. Annoyance is a linear function of time. So if a player waits t units of time they lose -nt satisfaction. They both still get m satisfaction if they meet. Note that the game is no longer symmetric.
Your job is still to find the Nash equilibrium wait times tb and tl. So you still have tofind the payoff functions of the players, the best response functions of the players and where these best response functions intersect, because that is the Nash equilibrium. This Nash equilibrium will depend on the parameters n and m, and so you will also have to describe how the equilibrium strategy profile changes with respect to these two parameters.
The old version of the problem will stand as a bonus assignment due at the end of the year so all those valiant efforts on that beast of a problem have not been in vain.
You may find the example we went over in class helpful.
Evolutionary Game Theory 
For the example in class I dont get the case of when your calculating the probability of case 1. you take the integral of 1/2 from ab+tb to 1, presumably to indicate the probability that lude arrives in this time segment. but then shouldnt the probability of byron, arrives and leaves before ludd be the integral from -1 to ab+tb?
Two things:
1) There is no class on the 15th as that is a holiday.
2) The link in the post is a write up of the exmaple from class or the assignment we are to do?
Sorry to be confusing Pat the post updated as I was editing it.
Why is this not a symmetrical game? I thought uniform distribution is always symmetrical…?
I wanted to say thanks for understanding the difficulties in this assignment and hooking us up. It wasn’t your obligation, and I’ve had enough professors completely ignore the fact work is a too complex. This simpler assignment has increased my understanding by 20 fold, and now I think I would be capable of the next one. But to sum it up, thank you!
Thank you for your understanding Dan!
What time are you holding the stats tutorial tomorrow?
The uniform probability distribution is a symmetric distribution, but the game is not symmetric because the players have different payoff functions. (One always arrives on time and the other arrives according to a probability distribution)
“Why is this not a symmetrical game? I thought uniform distribution is always symmetrical…?”
In addition to what Dan said, a game can only be considered symmetric if you can swap the roles of both players and the game would behave in exactly the same way, and governed by exactly the same payoff functions. This is not the case here because (as Dan said), their probability distributions are different.
How is this problem different from the example in class now that it’s also a uniform distribution?
It’s different in that only one of the players arrives with a uniform distribution. The other player always arrives exactly at time zero
oooo I see!
thanks a lot =)
Wah~~the 3rd edition is so easy~~~~~~~~~
Wah~the 3rd edition is so easy~~~~
“Half of the people can be part right all of the time, Some of the people can be all right part of the time, But all the people can’t be all right all the time. I think Abraham Lincoln said that.”
does Ludd know that Byron arrives at 0?
Yes Ludd knows that Byron arrives at 0. Also don’t be alarmed if certain symmetries pop up in the game. Do we move through the world or does the world move around us?
>< if Ludd knows Byron arrives at 0 then wouldn't Ludd just arrive at 0 too?
I'm confused.
so now we are not assuming that Tl and Tb are the same are we? ie. the wait time of the Byron and Ludd are now different?
“>< if Ludd knows Byron arrives at 0 then wouldn't Ludd just arrive at 0 too?
I'm confused."
Due to random circumstances, Ludd *can't* arrive at exactly t=0, although he tries. So, he has a probability of arriving between t=-1 and t=1.
"so now we are not assuming that Tl and Tb are the same are we? ie. the wait time of the Byron and Ludd are now different"
That assumption can't be made because the game is no longer symmetric. So, yes, they could be different.
…what do you mean Ludd “can’t” arrive at exactly t=0…
It’s different in that only one of the players arrives with a uniform distribution. “The other player always arrives exactly at time zero”
?
Yes, that is what I meant.
Byron always arrives at t=0.
Ludd can arrive at any time between t=-1 and t=1.
And hence, the problem. What is the optimum waiting time for Byron and Ludd so that they can maximize their payoffs.
Hope that helps.
does it mean that the integrals don’t involve ab any more since ab=0?
so we just set ab = 0?
“does it mean that the integrals don’t involve ab any more since ab=0?”
In a way, yes 😛
“so we just set ab = 0?”
Yes, because Byron always arrives at t=0
YAY
no more double integrals!
To add:
At least, that’s how I’ve done it. I’m getting a result that I’m not quite happy with. However, from a logical stand-point, the result makes sense.
I wonder if it’s one of Dan’s mean tricks…
hmm..it makes sense.. ab is 0..
nah.. Dan is not mean..
well.. setting ab to 0 i’m getting single integrals.. I’ll ask him tomorrow if this is right…
“well.. setting ab to 0 i’m getting single integrals”
Yes, setting ab to 0 will give you integrals (and you don’t even have to go as far as integrals, since the area can be easily calculated).
Note that getting single integrals is expected, since the probability is now based on *one* continuous random variable and not two as in the original question.
…. i got Byron’s payoff function as always concave up… does that mean something is wrong? since i’d always be solving for a minimum while we want a maximum -_-
arghhh
hmm..even tho it’s not symmetric, which means that t. and tb are not the same
the payoff functoins look the same.. except for the tb’s becoming tl’s…
“…. i got Byron’s payoff function as always concave up… does that mean something is wrong? since i’d always be solving for a minimum while we want a maximum -_-”
I seemed to get that too.
““…. i got Byron’s payoff function as always concave up… does that mean something is wrong? since i’d always be solving for a minimum while we want a maximum -_-”
I seemed to get that too.”
that’s a problem. because we want to find the max’s and then find the intersection. Now we only have the min, the max is infinity since it’s quadratic.
Maybe we should solve for Min and say what they shouldn’t do instead…
“the max is infinity since it’s quadratic.”
That’s not true for this situation, which is why I wasn’t concerned about getting that quadratic.
Do you know why?
““the max is infinity since it’s quadratic.”
That’s not true for this situation, which is why I wasn’t concerned about getting that quadratic.
Do you know why?”
… hmm… no….
or maybe because we are just looking for the intersection?
I’m really not sure.. argh
“… hmm… no….
or maybe because we are just looking for the intersection?
I’m really not sure.. argh”
Well here’s a hint:
What’s the maximum waiting time for Byron so as to ensure that Ludd is not going to arrive?
Similarly, what’s the maximum waiting time for Ludd?
What about the minimum waiting times?
“… hmm… no….
or maybe because we are just looking for the intersection?
I’m really not sure.. argh”
Well here’s a hint:
What’s the maximum waiting time for Byron so as to ensure that Ludd is not going to arrive?
Similarly, what’s the maximum waiting time for Ludd?
What about the minimum waiting times?
ohhh.. i think i get it now..we have an upper and lower bound..
but in this case.. wouldn’t we have 2 max’s as well instead of 1?
no you’ll on ly have 1 max…think again
Maybe this is just me but I am getting an erf function when i integrate the pdf. How on earth am I supposed to simplify this equation
“Maybe this is just me but I am getting an erf function when i integrate the pdf. How on earth am I supposed to simplify this equation”
You’ll get an erf if you try and integrate the normal distribution, because no such integral exists.
Daniel would like to add that: We are using uniform distributions now not normal distributions so hopefully you are not getting erf functions anymore.
“but in this case.. wouldn’t we have 2 max’s as well instead of 1?”
I got the same. I got 2 maxes, but each is dependant on the choice of m and n. I left my solutions completely algebraic.
i got 1 max though
Thanks Dan for understanding and scaling down the assignment, although we all might not admit it, we are grateful!
Have a great weekend.
I feel like too much knowledge about probability is being assumed for not having a Stats course as a prerequisite for this course.
Reply From Dan: I agree with this statement as applied to the first two iterations of the assignment, but now you only have to deal with a single probability distribution, the uniform probability distribution. Reading the first paragraphs of these two wikipedia entries may be helpful. Uniform distribution and Probability Density Function
Dan, I’m stuck…
To find the best response functions when there is no critical point that is a max, I check the end points right? But what are the end points? I’m taking two obvious end points for waiting time from just logically analyzing the scenario. And it seems that for reasonable values of n and m, we get that a particular end point always gives the larger value for the payoff function.
But then what would be my best response function? Would it just be the constant functions that take on the value of the end point giving the max payoff? If that’s the case wouldn’t Byron and Ludd always just wait that same amount of time to maximize their payoffs (unless n and m take on unreasonable values like if n >> m)?
Reply From Dan: So far it seems like you are on the right track. You have that the best response is always one of two end points, now you need to figure out which endpoints are the best response to eachother for which values of m and n. To figure out what the best responses actually is there is nothing for it but to evaluate the payoff function (not the derivative of the payoff funtion), at each possible combination of endpoints. There are two players and two endpoints so that gives you four points to check. If the payoff functions of the players are symmetric then it will be the case that Byron and Ludd always wait the same amount of time. You just need to specify which amount of time they will both be waiting for the various ratios of the values of m and n. I hope that helps.
Hint: There are three cases for the various values of m and n, one where they both wait one amount of time, one where they both wait the other, and one where it could go either way.
Just to clarify, are we using the normal probability function ( bell shaped) for ludd or are we assuming it is a straight line ( triangle shaped probability function with area of 1 under curve)
Reply From Dan: Actually as mentioned in this post we are using a uniform probability density function (rectangle shaped with area of 1 under the curve) since this is the easiest possible density function to deal with.
I am concerned with something both in the example and the assignment. In the example, the best response function was found to be a function of -c*(al-ab), which is the utility generated by waiting (al-ab) and to eliminate this we took the expected value of (al-ab) in this case and then subbed that in. However, shouldn’t we take the expected value of the utility function, not the utility function evaluated at the expected value? In the example, shouldn’t we have taken the double integral of c*(al-ab)*(distribution function) over possible values of al and ab?
Reply from Dan: This is a very good point Dave. You are absolutely right we should be taking the expected value of the utility function, and not the utility function evaluated at the expected value. Luckily for us taking bounded expectations is a linear operator and so in our case those two things happed to be the same thing. We are just skimming the surface of probability theory in this class and so I thought it best not to go into this detail, but I’m impressed with you for noticing.
So I’m kind of lost on how to find the probability of certain cases occurring. I don’t really understand how to make something like P(al + tl < 0) into integrals, or just how to figure out what that probability is. Anyone help?
Dan’s Reply: Remember that the probability of a Random variable taking on a value between two points is the Integral of it’s Probability density function integrated between those two points, or the area under the curve between those two points. In this case the only Random Variable you have to deal with is the arrival time of Ludd (a uniform random variable between -1 and 1, that means it’s probability density function is 1/2 on the interval -1 to 1 and zero everywhere else). Think of the four cases that can happen and then think about how they depend upon Ludd’s arrival time. I hope that is of some help.
“no you’ll on ly have 1 max…think again”
oooo! is it because the waiting times can only be positive, so we look at the *y* value for tb = 1 which is the max?
-_-
wait.. that doesn’t sound right T___T
HELP
where is dannn…ou ou ou ou ou ou est dan…does that mean the assignment is now due on friday ?:DDD
Dan’s Reply: I’m coming to the Math Help centre Now! (3:30) and will stay until just before 5
Question…
we know that E(al) = tb/2 for Byron’s case
but what about for Ludd? it seems more reasonable for it to be tl/2.. but should E(al) be the same?
stuck…
Dan’s Reply: It does seem more reasonable for it to be tl/2… and E(al) should be the same but what you actuall have is
E(al) given we are in the case where Byron arrives first and Ludd shows in time = tb/2
What case are you in when you need to know E(al) to compute Ludd’s payoff?
waiting at math help center …. don’t see Dan… =(
The case for Ludd’s payoff is Ludd arrives first and he needs to arrive within his own wait time tl, so that he’ll wait long enough for Byron to arrive at time 0.. he has to arrive within -tl to 0.. so E(al) should be tl/2.. is this wrong?
Looks right to me… and I am in the help centre, and will be until till 4:55pm. Don’t forget that it’s not actually E(al) (that is zero) but E(al) given a particular case.
Actually shouldn’t E(al) be -tl/2 (note the minus sign) not tl/2 ?
Yes, but the average time Ludd waits is 0 – E(al), (o being Byron’s arrival time) so it all comes out in the wash. I was just jumping the gun.
Someone sent me this:
Just wondering if we’re still assuming m< (n/2) in the scaled back version of assignment #3.
I replied:
No we are not. We are letting m and n be in all sorts of ratios to each other. In fact you have to figure out what the different Nash equilibria are for all possible ratios of m and n. (We are still assuming that both are positive)
-daniel