Hedging on a weather prediction

[Hill]

Homework Statement

Weather forecaster Ra is so certain it will rain on Sunday that he would offer a bet of 3:2 that it will happen. On a different channel, Su is confident that on Sunday it will not rain; she offers 4:3 odds.
(a) A person has $100. Set up a hedge where the person will win the same amount of money no matter what happens on Sunday.

Relevant Equations

earnings=gain-loss

Let the person bet $x$ with Ra and $100-x$ with Su. If it rains on Sunday, the earnings are $\frac 4 3 (100-x)-x$. If it's sunny, the earnings are $\frac 3 2 x - (100-x)$.
$\frac 4 3 (100-x)-x=\frac 3 2 x - (100-x) \Rightarrow x \approx 48$.
I'd like to get a confirmation that my understanding is correct.

 

[PeroK]

It's late and I'm tired, but my understanding is:

You bet $x$ with Ra that it will not rain. If it rains, you lose $x$. If it does not rain, then you receive $\frac 5 2 x$. That's odds of 3:2 plus your original stake.

You bet $1 -x$ with Su that it rains. If it rains, you receive $\frac 7 3 (1 - x)$. If it does not rain, you lose $1 - x$.

You then equate the amount of money you have in the two cases:
$$\frac 5 2 x = \frac 7 3 (1 - x)$$Which gives $x = \frac {14}{29}$. And, you have $\frac{35}{29}$, which means you win $\frac{6}{29} \approx \$20.69$.

 

[PeroK]

PS I see now this is the same as your answer, more or less: $x = \frac {14}{29} \approx \$48.28$.

 

[Hill]

The second part of the question:

Both forecasters believe they are making a fair bet; that is, they are willing to take either side. Determine the likelihood each has for rain on Sunday, for no rain on Sunday.

Let $P_F(w)$ be a likelihood that the forecaster $F$ has for the weather event $w$.

Then, if it's a fair bet,

$P_{Ra}(rain)=\frac 3 2 P_{Ra}(sun)$

and

$P_{Su}(sun)=\frac 4 3 P_{Su}(rain)$.

Considering that

$P_{Ra}(rain) +P_{Ra}(sun)=1$

and

$P_{Su}(sun)+P_{Su}(rain)=1$

I get

$P_{Ra}(rain)=\frac 3 5, P_{Ra}(sun)=\frac 2 5$

and

$P_{Su}(sun)=\frac 4 7, P_{Su}(rain)=\frac 3 7$.

 

[Hill]

A similar exercise:

Two people have different opinions on the Tennessee and Florida game to be played tonight. One person supports Tennessee and is willing to offer 11 to 10 odds; the other person supports Florida and is willing to offer 5 to 3 odds. Go through the analysis to determine how much to bet with each person. If Heili has $100 to invest, how much is she assured of earning?

Let $x$ be a bet with person 1, and $1-x$ a bet with person 2.

If TN wins the game, the earnings are $\frac 5 3 (1-x) - x$.

If FL wins the game, the earnings are $\frac {11} {10} x - (1-x)$.

Equating

$\frac 5 3 (1-x) - x=\frac {11} {10} x - (1-x)$

I get $x=\frac {80} {143}$ which makes the assured earning $0.175$. Thus, on $100 Heili is assured to earn $17.5.

(Unless I've made a bad arithmetic error.)

 

[pbuk]

Let the person bet $x$ with Ra and $100-x$ with Su.

That is not the correct approach. Instead, Let the person bet $r$ with Ra and $s$ with Su. You should be able to determine the ratio $\frac r s$ which will enable you to easily write down example values of R and S in dollars to give a perfect hedge (although the correct term in this case would be a perfect arbitrage).

 

[pbuk]

If that is the whole statement of the question, I think you have also misunderstood what it means [EDIT: although that doesn't make any difference to the answer in this case because of the symmetry].

Weather forecaster Ra is so certain it will rain on Sunday that he would offer a bet of 3:2 that it will happen.

Ra offers a bet of 3:2 that it will rain. So if you place this bet and it rains then you win; if it does not rain you lose.

Su is confident that on Sunday it will not rain; she offers 4:3 odds .

Su offers 4:3 odds that it will not rain. So if you place this bet and it does not rain then you win; if it rains you lose.

 

[pbuk]

Note that the question is:

Set up a hedge where the person will win the same amount of money no matter what happens on Sunday.

so the answer is in the form "I bet $r with Ra and $s with Su".

Neither "I bet approximately $48 with Ra and apprixmately $52 with Su" nor "I bet $\frac{14}{29}$ with Ra and $\frac{15}{29}$ with Su" are correct answers.

 

[PeroK]

Note that the question is:



so the answer is in the form "I bet $r with Ra and $s with Su".

Neither "I bet approximately $48 with Ra and apprixmately $52 with Su" nor "I bet $\frac{14}{29}$ with Ra and $\frac{15}{29}$ with Su" are correct answers.

I assumed that Ra wasn't accepting bets that it rains, but rather accepting bets that it would not rain.

And that the odds of 3:2 apply to a bet that it does not rain. The odds might not be 2:3 on a bet that it rains. That would only be the case if they were offering fair odds on both outcomes.

 

[Hill]

If that is the whole statement of the question...

You are right, this is not the whole statement of the question, I've skipped the following note (I'm sorry!):

Remember, if it rains after someone bets with Ra, Ra keeps all of the money. If it does not rain, then Ra pays the better $3 for each $2 that was wagered.

 

[PeroK]

As an aside, a friend of mine said there's a guy in Finland who trawls the Internet looking for such betting anomalies. And makes a good living from it.

The betting sites that I've looked at occasionally in this context have different odds for the same outcome. But, not different enough to allow a guaranteed win. They must try to avoid such anomalies somehow, I imagine.

 

[Gavran]

The second part of the question:



Let $P_F(w)$ be a likelihood that the forecaster $F$ has for the weather event $w$.

Then, if it's a fair bet,

$P_{Ra}(rain)=\frac 3 2 P_{Ra}(sun)$

and

$P_{Su}(sun)=\frac 4 3 P_{Su}(rain)$.

Considering that

$P_{Ra}(rain) +P_{Ra}(sun)=1$

and

$P_{Su}(sun)+P_{Su}(rain)=1$

I get

$P_{Ra}(rain)=\frac 3 5, P_{Ra}(sun)=\frac 2 5$

and

$P_{Su}(sun)=\frac 4 7, P_{Su}(rain)=\frac 3 7$.

Your results are correct.
The expected value of a profit (or a loss) in the case of a fair bet must be equal to zero.

$E(\text{Profit}_{Ra})=3\cdot P_{Ra}(sun)-2\cdot P_{Ra}(rain)$
$E(\text{Profit}_{Ra})=3\cdot 2/5-2\cdot 3/5$
$E(\text{Profit}_{Ra})=0$

$E(\text{Profit}_{Su})=4\cdot P_{Su}(rain)-3\cdot P_{Su}(sun)$
$E(\text{Profit}_{Su})=4\cdot 3/7-3\cdot 4/7$
$E(\text{Profit}_{Su})=0$