Math Challenged

[tdvance]

Let X be uniformly distributed on [-1,1]. Let Y=X^2.

Then, they are not independent:
probability that X>0.9 and Y<0.5 is 0, but prob X>0.9 is 0.1, prob Y<0.5 is something close to .7 or so, product is .07 != 0.

However, they are uncorrelated (i.e. have 0 covariance):

COV(X,Y) = E((X - E(X))(Y-E(Y)) = E((X - 0)(X^2 - 1/2))

= E(X^3 - 1/2X) = 0.

Todd

[hhEb09'1]

Let X be uniformly distributed

In the problem, both X and Y are normally distributed

[tdvance]

oops--I forgot about that.

Todd

[The_Radiation_Specialist]

Did this game die?

[Sarawak]

I think I killed it

Maybe someone else should ask a new question.

[The_Radiation_Specialist]

Since hhEb09'1 answered my question correctly, I think he may go now.

[hhEb09'1]

An oldie but a goodie. And the eighth graders were working on it last month

Take a sphere, and drill an eight inch long (or 8 cm long, if you prefer) hole through it. What is the volume of the material that is left in the sphere?

[The_Radiation_Specialist]

I remember this...

Ah yes, the mighty Google gives me the thread...

http://www.bautforum.com/off-topic-b...nt-people.html

So that is 4*4*4*4/3*pi = (85 + 1/3)*pi cubic inch (or cm)

[hhEb09'1]

the thread...

http://www.bautforum.com/off-topic-b...nt-people.html

I was sick that week

Yep, that is the answer (and the discussion), you're up!

[The_Radiation_Specialist]

Prove that

9a⁴-8a³b+b⁴ ≥ 0

holds for all real numbers a and b.

Hint: use the AM GM Inequality

[Sarawak]

Probably there is a more elegant solution, but:

9a⁴-8a³b+b⁴ = (b-a/sqrt(2)(i-sqrt(1-4isqrt(2))))*(b-a/sqrt(2)(i+sqrt(1-4isqrt(2))))*(b-a/sqrt(2)(-i-sqrt(1+4isqrt(2))))*(b-a/sqrt(2)(-i+sqrt(1+4isqrt(2))))

For a=0, the function is ≥0 because it is b⁴. For real a<>0, the function is continuous and all its zeros (in b) have non-zero imaginary part, so the function cannot have both positive and negative values. And for b=0, it is 9a⁴, so for real a<>0 and real b it is always positive.

[The_Radiation_Specialist]

Probably there is a more elegant solution

The other solution is we divide it into 4 parts ... 9a^4 + b^4 = 4a^4 + 4a^4 + a^4 + b^4
and take the AM of LHS and GM of RHS and according to the inequality we can put the sign ... (9a^4 + b^4)/4 ≥ 4th root(4a^4 * 4a^4 * a^4 * b^4 )
so ... 9a^4-8a^4b+b^4 ≥ 0

Your go.

[Sarawak]

My last turn was not a big success, I will try to do better this time. First I will give a solution to my earlier question.

Choose X as a normal random variable, and Z with values +1 or -1 with equal probability, and Z is independent of X. Then Y=Z*X is a normal random variable, it is uncorrelated with X, but it is not independent of X.

Now here is my new question.

In a gambling game, a gambler wins 0.5 for each 1.0 of the bet if a coin toss is "heads," and loses 0.4 if the coin toss is "tails." So if he bets five Australian dollars, then he wins 2.5 if the coin is heads, and loses 2.0 if it is tails. Now, suppose someone plays this game repeatedly with the following strategy. He begins with a gamble of one Australian dollar. Whatever he has after this gamble (1.5 or 0.6), he bets the entire amount on a second coin toss. After this, he can have 2.25, 0.9, or 0.36, but he then bets the full amount on a third coin toss. And he does this repeatedly until N coin tosses are complete.

What is his average winning (including the one Australian dollar he began with) after N coin tosses? As N goes to infinity, what is his average winning?

What is his median winning (including the one Australian dollar he began with) after N coin tosses? As N goes to infinity, what is his median winning?

[hhEb09'1]

In a gambling game, a gambler wins 0.5 for each 1.0 of the bet if a coin toss is "heads," and loses 0.4 if the coin toss is "tails." So if he bets five Australian dollars, then he wins 2.5 if the coin is heads, and loses 2.0 if it is tails.

There is a different way of representing the game. I would have said he starts with a stake of five dollars, bets 40% of it (2 dollars) and has good odds of winning (he wins $2.50 half of the time on a $2 bet.)

[Sarawak]

There is a different way of representing the game. I would have said he starts with a stake of five dollars, bets 40% of it (2 dollars) and has good odds of winning (he wins $2.50 half of the time on a $2 bet.)

Yes, this is an equivalent formulation. So if anyone prefers to think in this way, he bets 40% of his money on each coin toss. If he loses, all the amount bet is lost, but if he wins, he receives back 1.25 times the bet, plus the original bet. So if he has $10, he bets $4. If he loses, the $4 is gone, so he has $6 remaining, and his next bet will be $2.40. If he wins, he receives back his $4 plus $5, so he has $15 in total, and the next bet will be $6.

Also, for the median, it is simpler to think only of the case where N is even. The case where N is odd has some slight complication, and does not add anything interesting to the problem.

[Sarawak]

I hope I have not killed this thread again

Someone please try, it is not so difficult. If no one can answer, I will give some hints. Here is the first hint. For the median case, it can be helpful to consider the logarithm of is winnings instead of winning. When N is even, the median of the logarithm is the logarithm of the median.

[hhEb09'1]

Then Y=Z*X is a normal random variable, it is uncorrelated with X, but it is not independent of X.

What do you mean by Z*X?

[Sarawak]

What do you mean by Z*X?

It is just ordinary multiplication, not convolution or anything strange. Z has values +1 or -1 with equal probability, and X has a random value with the normal distribution. If X and Z are independent, then the product of the two random variables is also a normal distribution, and it is uncorrelated with X. But it is not independent.

[The_Radiation_Specialist]

I got some help from a friend, will that count?

notice that if the gambler wins with a bet x, his total amount of money becomes 1.5x. If he loses, then it becomes 0.6x. So if there are N tosses, the money he gets, y(x), is so that y(x) = x*1.5^a * 0.6^b where a + b = N. So a = 0,1,2, 3 ... N and b = N, N-1, N-2...0. So there are N + 1 ways of having a + b = N. That said, we look at 1.5^0*0.6^N + 1.5^1*0.6^N-1 + ... + 1.5^N*0.6^0. This is a geometric sum of N +1 terms with the first term equal to 0.6^N and the argument being 1.5/0.6. So that sum is equal to 0.6^N * ((1.5/0.6)^N+1 - 1)/(1.5/0.6 -1). This really simplifies to 5/3 * 1.5^N - 2/3 *0.6^N. Taking the average, we get 5/3 * 1.5^N - 2/3 *0.6^N / N + 1. From this it's clear that the average as N goes to infinity is infinity: 0.6^N / N + 1 obviously goes to 0, and 1.5^N / (N + 1) goes to infinity since 1.5 > 1 and the order of magnitude of e^x is greater than x + 1.

[The_Radiation_Specialist]

There is something wrong with this thing, it won't let me post the entire proof in one post.

Now considering the median: from the geometric sum above, we have an increasing sequence of N + 1 terms. So if N is even, the median is 0.6^N * 1.5^(N/2 + 1) = . If N is odd, the median is the average of 0.6^N * 1.5^(N+1 /2) and 0.6^N * 1.5^(N+1 / 2 + 1). But in all cases the we get a development that yields numbers lesser than 1 to unbounded powers. So the median goes to 0.

[Sarawak]

The Radiation Specialist, I think you have the right idea, but there are some details. There are N+1 ways to have a+b=N, but they do not all have equal probability. To have a=b=N/2 is much more probable than to have a=0 and b=N. The probability of a=M and b=N-M is {(N)!/[(M)!(N-M)!]}/(0.5^N), so to find the mean you need to weight by these probabilities, not give equal weight to all values of M as you did.

But I think an easier way to find the mean is to call X(i) the multiplier of gamble i, so X(i) is 1.5 if he wins, and X(i) is 0.6 if he loses. Then he has X(1)*X(2)*...*X(N-1)*X(N). The problem is to find the mean of this quantity. But if random quantities are independent, then the mean of their product is the product of their means. And the mean of X(i) is the same for all values of i, so the mean of his winnings after N gambles is some number raised to the power of N.

For the median, I think you have the right idea but just made a small typo. The median outcome for even N has N/2 wins and N/2 losses, so the median amount of winnings is 0.6^(N/2)*1.5^(N/2).

You are very close, I think you can find an exact expression for the mean if you do it in this way

[The_Radiation_Specialist]

Prove that the fraction,

21n+4
-------
14n+3

is irreducible for every natural number n

[tdvance]

Prove that the fraction,

21n+4
-------
14n+3

is irreducible for every natural number n

Let (.,.) represent GCD. Because n is a natural number, the arguments to GCD below are all integers, and, just writing out Euclid's algorithm by hand:

(21n+4,14n+3) = (21n+4 - (14n+3),14n+3) = (7n+1,14n+3)
= (7n+1,14n+3-2*(7n+1)) = (7n+1,1) = 1, so the numerator and denominator have no common factor (other than 1 or -1) and the fraction is irreducible.

[The_Radiation_Specialist]

Yes, your turn.

[Sarawak]

I want to say a few things about the solution to my question. The mean is 1.05^N, and the median is 0.9^N/2. It is true also for any percentile that it approaches zero as N goes to infinity. So the gambler following this strategy becomes rich on average, but becomes nearly broke (you can define "nearly broke" to be as close to zero as you like) with probability that approaches one.

[tdvance]

For n a nonnegative integer, what (in terms of n) is the sum of the n-th powers of the roots of the polynomial x^6 - 1 ?

[The_Radiation_Specialist]

1+(-1)^n

[Sarawak]

If you mean to include all complex roots in addition to real roots, then the sum is 6 if n is divisible by 6, and 0 otherwise. But if this is the correct answer, then I give my turn to The Radiation Specialist, because my questions have not been very successful.

[tdvance]

Homo Bibiens has the correct answer, but has given his turn to The Radiation Specialist.

Todd

[The_Radiation_Specialist]

Show that there are infinitely many pairs of positive integers (m, n) such that

(m + 1) / n + (n + 1) / m

is a positive integer.



I'll be away for a few days so if anyone thinks they got it right they may go ahead with the next question.