Expected Value of Election Results

[vparam]

Homework Statement

Suppose that in the 2050 election, polls show Party A has a 60% chance of winning the senate and a 50% chance of winning the house, and that if Party A wins the house, it has an 80% chance of winning the senate. Let X be the random variable whose value is the the number of congressional houses won by Party A. (a) Find the expected value of X.

Relevant Equations

$p(S|H) = \frac{p(S \cap H)}{p(H)}$
$E(X)=\sum_{s \in S}{p(s)X(s)}$

I submitted this solution, and it was marked incorrect. Could I get some feedback on where I went wrong?

Let S represent the event that Party A wins the senate and H represent the event that Party A wins the house.

There are 4 cases: winning the senate and house ($S \cap H$), winning just the senate ($S \cap \overline{H}$), winning just the house ($H \cap \overline{S}$), and winning neither ($\overline{S} \cap \overline{H}$).

I computed the probabilities for each compound event as follows:
$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4$$ $$p(S \cap \overline{H}) = p(S)p(\overline{H}) = (.6)(.5) = .3$$ $$p(H \cap \overline{S})=(p(S) - p(S \cap H))p(H) = (.6 - .4)(.5) = .1$$ $$p(\overline{S} \cap \overline{H}) = p(\overline{S})p(\overline{H}) = (.4)(.5) = .2$$
This means that the expected value is:
$$E(X) = 2(.4) + 1(.3) + 1(.1) + 0(.2) = 1.2$$

 

[PeroK]

There are 4 cases: winning the senate and house ($S \cap H$), winning just the senate ($S \cap \overline{H}$), winning just the house ($H \cap \overline{S}$), and winning neither ($\overline{S} \cap \overline{H}$).

Your overbars seem to show up when I quote your post, but I can't see them in the original.

I computed the probabilities for each compound event as follows:
$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4$$ $$p(S \cap \overline{H}) = p(S)p(\overline{H}) = (.6)(.5) = .3$$ $$p(H \cap \overline{S})=(p(S) - p(S \cap H))p(H) = (.6 - .4)(.5) = .1$$ $$p(\overline{S} \cap \overline{H}) = p(\overline{S})p(\overline{H}) = (.4)(.5) = .2$$

You have a total probability of winning the Senate as $0.7$. And, you've assumed the probabilities are indepenent in some of those calculations. I drew a Venn diagram.

This means that the expected value is:
$$E(X) = 2(.4) + 1(.3) + 1(.1) + 0(.2) = 1.2$$

Two of those numbers are wrong.

 

[FactChecker]

$$p(S \cap \overline{H}) = p(S)p(\overline{H}) = (.6)(.5) = .3$$

Only if S and H are independent events. But it is easy to show that they are not independent.

$$p(\overline{S} \cap \overline{H}) = p(\overline{S})p(\overline{H}) = (.4)(.5) = .2$$

Again, only if S and H are independent events.

 

[vparam]

I drew a Venn diagram.

Thank you for the reminder to use a Venn diagram! The problem breaks down more easily after drawing one out.

I've redone the calculations as follows:
$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4$$ $$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$ $$p(H \cap \overline{S}) = p(H) - p(S \cap H) = .5 - .4 = .1$$ $$p(\overline{S} \cap \overline{H}) = 1 - p(S \cup H) = 1 - (p(S) + p(H) - p(S \cap H)) = 1 - (.6 + .5 - .4) = .3$$
This means that the expected value is:
$$E(X) = 2(.4) + 1(.1) + 1(.2) + 0(.3) = 1.1$$

Only if S and H are independent events. But it is easy to show that they are not independent.

Again, only if S and H are independent events.

And thank you for pointing this out. This explains why my previous calculations don't make sense.

Your overbars seem to show up when I quote your post, but I can't see them in the original.

Is \overline the standard code for overbars, or should I use something else?

Thank you both for the help!

 

[PeroK]

Thank you for the reminder to use a Venn diagram! The problem breaks down more easily after drawing one out.

I've redone the calculations as follows:
$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4$$ $$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$ $$p(H \cap \overline{S}) = p(H) - p(S \cap H) = .5 - .4 = .1$$ $$p(\overline{S} \cap \overline{H}) = 1 - p(S \cup H) = 1 - (p(S) + p(H) - p(S \cap H)) = 1 - (.6 + .5 - .4) = .3$$
This means that the expected value is:
$$E(X) = 2(.4) + 1(.1) + 1(.2) + 0(.3) = 1.1$$

And thank you for pointing this out. This explains why my previous calculations don't make sense.Is \overline the standard code for overbars, or should I use something else?

Thank you both for the help!

I don't understand why the Latex renders only when I quote your post and not in the original. I can't see the overbars in your posts, but I can when I reply.

PS it doesn't work when I take them out of the quote box!

$$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$

 

[vparam]

I don't understand why the Latex renders only when I quote your post and not in the original. I can't see the overbars in your posts, but I can when I reply.

PS it doesn't work when I take them out of the quote box!

$$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$

I'm not sure why that's the case since I'm very new to Latex and it seems to show up properly on my end. I wrote up everything again using an alternate notation instead for visibility.

Probability calculations:
$$p(S \cap H) = p(S | H) p(H)= (.8)(.5) = .4$$ $$p(S \cap H^{\mathrm{C}}) = p(S) - p(S \cap H) = .6 - .4 = .2$$ $$p(H \cap S^{\mathrm{C}}) = p(H) - p(S \cap H) = .5 - .4 = .1$$ $$p(S^{\mathrm{C}} \cap H^{\mathrm{C}}) = 1 - p(S \cup H) = 1 - (p(S) + p(H) - p(S \cap H)) = 1 - (.6 + .5 - .4) = .3$$

Expected value computation:
$$E(X) = 2(.4) + 1(.1) + 1(.2) + 0(.3) = 1.1$$

 

[Office_Shredder]

1.1 is right. Good job working out the probabilities. Also, his is a trick question!

Expectancies add linearly. Let $S$ be the number of senate seats won, and $H$ the number of house seats won (each is either 0 or 1. This is called an indicator variable and is a very common idea to use)

Then you want to compute $E(S+H)=E(S)+E(H)$. The correlations are irrelevant.

 

[FactChecker]

I don't understand why the Latex renders only when I quote your post and not in the original. I can't see the overbars in your posts,

Same here.

but I can when I reply.

PS it doesn't work when I take them out of the quote box!

$$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$

When I reply to you, the overline of the first H shows.

$$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$
And now it doesn't. (There is an overline of the first H. If it matters, I am using the Firefox browser.)