Game Theory Question

[HP]

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How you figured it out.

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thought so just wanted to make sure so i don't waste any effort. this will take me a while to write out, in the middle of it now

[HP]

Okay, so say our opposing firm's probability distribution is g(x)

Also, let's assume for now that there is no 'pure strategy' components in g(x) (this intuitively makes sense for the reason you mentioned before). Let's assume it's a nice looking probability density function. In other words:



Okay so say we choose a price P. P can be between 0 and 1. Our EV then is:



now, for the opposing firm to be using a strategy from a Nash Equilibrium thing, the following must be true:



where C is some constant. If this were not the case, then a certain value of P would would become the best choice for us, and hence our opposition's strategy is exploitable, and not part of a Nash Equilibrium thing

Also it should be noted our EV could actually be less than C, if we choose a P such that:



for any possible nonzero delta. But let's not worry about that just yet

okay so back to this equation:



with a bit of alegbra:



and now if we take the negative derivative of both sides with respect P:



the left side reduces to just g(P). Differentiating the right side, we get:



Now, we know g(x) is 0 when x>1. Intuitively, we want to be able to choose as high a P as we can. So we can assume that g(x) will have the form:

C/(q*x^2) when x<1

0 when L<x

Now, if we know what C is, we know L, or vica versa because the integral of the probability distribution must equal 1

You can show mathematically if you choose a P > L, your EV will equal C. However what happens if you choose a P <= L, trying to undercut his whole range?

If we undercut the competitor's whole range, we can not have an EV greater than C (as this would imply the competitor's strategy is not part of a Nash Equilibrium thing)

If your strategy is to undercut the competitor, you should set your price at P=L. This will maximize your EV.

Well, if we choose L, our EV is simply L (since we will always have the lowest price)

so, L has to be less than or equal to C (or else the competitor's strategy is not a Nash Equilibrium thing). Our competitor wants an L as high as possible (because this means he gets to choose a higher price on average and make more money from the consumer) so let's set L=C

and now, because the integral of the probability distribution has to equal one, the following must be true:



solving this for L we find that L equals (1-q)

so finally, this gives us g(x):



I think there may be some minor logical holes in my method (I'm not strong in math), however I think my answer is correct in hindsight, so you might philosophically say my process was a glorified guess and check

[janneman]

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now, for the opposing firm to be using a strategy from a Nash Equilibrium thing, the following must be true:


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Upto this point your argument is right i think (for finding the Nash equilibrium of the game at least, i m not sure that is the question though)

However here you make a mistake, the lefthandside does not have to be constant over the whole range, just locally at our P so either P=0, P=1 or d/dP of the lefthandside is zero, but only at our P. If the lefthandside was really constant it would be zero by taking P=0.

[HP]

I believe this is covered with this part of my post:

Also it should be noted our EV could actually be less than C, if we choose a P such that:



for any possible nonzero delta.

[HP]

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so finally, this gives us g(x):



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err, that should be a '1-q' there for the range of x

(1-q)<x<1

[tabako]

Thanks

[janneman]

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I believe this is covered with this part of my post:

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It clearly isn't. First of all first you define C to be the E.V. (and give the right definition). Then you say our E.V. can be less (not true) if your integral is zero for all delta. If P=0 your integral is 1 for delta=1 so C is still zero if it is constant.

Notice that this is not just a formality, your computation depends on C being constant which it clearly is not.

I think your calculation (when done right) will give us the optimal P for any distribution g. However it will not give g.

[HP]

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It clearly isn't. First of all first you define C to be the E.V. (and give the right definition). Then you say our E.V. can be less (not true) if your integral is zero for all delta. If P=0 your integral is 1 for delta=1 so C is still zero if it is constant.

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Like I said I'm not good at math, and a miscommunication may have arisen from me writing down the wrong thing

In the P = 0 case, the integral can be zero, given a small enough delta. What I was trying to say, is if it can possibly be zero, then it's possible your EV is less than C

This is because you've chosen a P that your opponent would never have chosen. Maybe I should have just said that.

If you choose a P that your opponent would never have chosen, it means you are choosing a P that is not part of a Nash Equilibrium strategy, hence you are not guaranteed to have an EV of C

[HP]

btw, I can tighten my restrictions further. If the following can be 0 given the right choice of a non-zero delta:



then P is not part of a Nash Equilibrium strategy, hence you are not guaranteed an EV of C

This is to avoid confusion in the case P is exactly on the edge of the domain of g(x)