Ok, thank you very much.

ty pair the board. So any equation that has a square root in it cannot be a function since there will always be two outputs for x??

PTB (or anyone else),

I wasn't aware that risk aversion could be quantified. Could anyone explain what he's referring to here? Thanks.

What is the difference between angular speed and linear speed??

An equation is not a function. An equation in two variables x,y may be used to define a function f(x) in the x variable according to the (x,y) solutions to the equation iff there are no solutions with the same x value having different y values.


There were no square roots in the equation you presented.

y - x^2 = 0

defines a perfectly good function, f(x).

It also depends on which variable the function is to be defined.

x - y^2 = 0

Does not define a function in the x variable but does define a function in the y variable.


PairTheBoard

Investor A needs an extra 8% return to be willing to accept the risk associated with an increase in s.d from 15% to 20%, whereas investor B is willing to take that extra risk for only a 3% increase in return - so A is more risk averse.

You could try and quantify risk aversion - it should have something to do with how concave their utility curve of money is. The standard measure of risk aversion is to suppose that your utility curve is a function u(w), then your risk aversion is -u''(w) / u'(w), the Pratt-arrow measure

There is a wikipedia page with more info, http://en.wikipedia.org/wiki/Risk_aversion.

This is mistaken.

For the first one, use tan(x) = sin(x)/cos(x) -substitute this into the LHS. You can the use the fact sin^2(x)+cos^2(x)=1 to simplify the result. Then using tan=sin/cos again you get to the RHS.

The second reduces to showing that tan^2x+1/tan(x)=1/(cos(x)sin(x)). You can use the same techniques as above to do this.

little rusty on this stuff, its been a while since i took my last probability class but iirc the functions g(x) and h(y) can be any combination of functions as long as they're factors of the joint pdf

so you could look at something like

f(x,y) = 4xy 0 < x < 1 and 0 < y < 1

the two marginal pdfs are f₁(x) = 2x and f₂(y) = 2y...you can let g(x) = 4x, h(y) = y or g(x) = x, h(y) = 4y and so on

easy way to quickly determine if two random variables are independent i was taught, the following theorem is directly from an old textbook:
Two random variables X and Y with joint pdf f(x,y) are independent if and only if:

• the support set {(x, y)|f(x,y) > 0}, i s a Cartesian product

• the joint pdf can be factored into the product of functions of x and y, f(x,y) = g(x)h(y)

both of these must be met...if you have a joint pdf like

f(x,y) = x + y when 0 < x < 1 and 0 < y < 1

then the first requirement is met, the support set is a cartesian product but the second requirement is not met, f(x,y) isn't a product so it can't be factored

now take a pdf like

f(x,y) = 4xy when 0 < x < y < 1 and zero at all other points

you can just look at this and see that the support set is not a Cartesian product, it is a triangle region so right away you know that the variables are not independent

stuck on system dynamics, group project and we're all struggling to get anywhere
i'm not looking for the answer, just a nudge in the right direction since we're having trouble getting started

the red bolded part isn't for emphasis, there was a typo on a previous version which was corrected there

hmm not sure, think he´s referring to a slope which measures how risky portfolios are, like (expected return - risk free) / standard deviation and the higher the line is, the more suitable it is for a risk averse investor.

The question I posted is supposed to be even simpler I believe, someone else suggested B since he chooses a lower safe return rather than a higher unsafe return.

Anyways, thanks for the input guys. I appreciate the time.

Hi guys.

So last night I got an email from a couple of friends working on something at the university. They asked me for help on some math. It's been a couple of years since I took classes in this stuff so I'm, to put it mildly, very rusty.

Anyway, they asked me if I could determine whether or not the function Q_n is pseudoconcave.


note: in the second line with the sum, it's the sum over the whole expression, so there should probably have been a parenthesis. Furthermore W is a positive definite matrix. " ' " denotes the transpose.


The first thing I did was to check up on all the definition (it was needed!), ie positive definiteness, conditional expectation and especially pseudoconcave functions. The definition on pseudoconvex functions on wiki (http://en.wikipedia.org/wiki/Pseudoconvex_function), which redirects to "Plurisubharmonic function" which I haven't heard of before, doesn't seem all that helpful so I found my old textbook, which give me following definition of a pseudoconvex function:
--
f is pseudoconvex if
for each x1, x2 with grad(f(x1))'(x2-x1) => 0 we have f(x2) >= f(x1)
or equivalently, if f(x2) < f(x1), then grad(f(x1))'(x2-x1) < 0.

and f is pseudoconcave if -f is pseudoconvex.
--

My plan was to reverse all the above inequalities (obtaining a definition of a pseudoconcave function) and then check the function Q_n and see if it satifies it.
I tried playing around with it a bit but with no succes. Right now I (amongst other things) fail to see where the conditional expectation plays in. It would be obvious to try and find the gradient of Q_n, but it seems the expression will get pretty ugly so I'm not sure it's the right thing to do.


Anybody have any suggestions or anything that can get me started?



PS. Just came to think of that convexity (concavity) implies pseudoconvexity (pseudoconcavity) and it might be easier to show that the function is concave. I'm gonna look into that next.

PPS. (very) dumb question: There's a difference between these two functions, right?
f(x,y), where x denotes the first coordinate and y the second, and
f(x), where x is some vector, x = (x, y)'.
If yes, how do I differentiate the latter, can I just treat x as onedimensional, ie just find f'(x) as if x where in R?

Reviewing some thermodynamics and I remember seeing in one equation work was stated as

w= pdv + vdp , and I can't for the life of me remember when in the hell this equation would apply?

Can anything else other than an emf induce current in a solenoid?

What can self inductance specifically be used for?

Any easy expermints that can be use to demonstrate the principles of self induction?

All help much appreciated!

FML, Philosophy final exam in 45 minutes ... have to write an in-class essay on one of four topics (he gave us the four, didn't tell us which one he would pick). Has to do with the "future like ours" argument, Ayer, Kant / Mill, or Hobbes.

I suck at induction part 2:

To do a proof by induction I have to have the one-case be true, then assume the general case is true, and prove that the successive case holds. But somehow I'm just awful at this:

want to prove the associative law for addition

(m+n) + k = m + (n+k)

one-case: (m+n) + 1 = m + (n+1) (are we just assuming this is true?)

So, we suppose (m+n) + k = m + (n+k) and must show that (m+n) + (k+1) = m + [n+(k+1)].

But the text starts with: (m+n) + (k+1) = [(m+n) + k] + 1 (by the one-case)...I'm not sure what they mean here by "by the one case"...

Sorry if this is a super basic question but I'm obviously missing something.

Thanks for the help guys,
Mariogs

Well, I'm guessing you have to "prove" this in some way - but I don't really see how, since I'm not sure what axioms you're allowed to use.
Let's assume you've proven the "one case" k=1,
(m+n) + 1 = m + (n+1) [1]
Then the equation we're starting with
(m+n) + (k+1) = [(m+n) + k] + 1
can be written
u + (v+1) = (u+v) + 1
with u=m+n, v=k. And we know this is true from [1].

You have to start your theorem more carefully (I don't even know what setting you're in: R, N, Z, ???). So, for instance, we could say:

Thm: For all natural numbers m, n, k, we have that (m+n) + k = m + (n+k).

Proof: By induction on k. Fix m and n, natural numbers. Let K be the set
{k in N : (m+n) + k = m + (n+k)}. In this proof, we will use two axioms of natural numbers.
(A1): for any a in N, a + 0 = a
(A2): for any a, b in N, (a+b)+1 = a+(b+1)

Now, to the proof. Notice that 0 is in K by (A1); that is, (m+n)+0 = m+n = m + (n+0). Now, since K is nonempty, choose k from K. We'll show that k+1 is also in K.

(m+n) + (k+1) = ((m+n)+k) + 1, by (A2)
....................= (m + (n+k)) + 1 since k is in K.
....................= m + ((n+k) + 1) by (A2)
....................= m + (n + (k+1)) by (A2).

So k+1 is in K. By PMI, then, K = N, and the theorem follows.

Edit: So let's explain induction a bit more carefully. The principle of mathematical induction says that if we have a property that is either true or false for all natural numbers (call it P(n): so maybe P(0) is true, P(2) is false, etc. In our case above, P(k) is true if (m+n)+k = m+(n+k), and we want to show P(k) true for all k in N), then
if P(0) is true AND [ P(k) true => P(k+1) true], then P(k) is true for all k.

You might view induction as dominoes. If you knock down the k_th domino, the (k+1)st one falls. So as long ass you knock down 1 domino, all of the succeeding ones fall. i.e. if we can knock down P(0), and if [ P(k) true => P(k+1) true] (i.e. if knocking down P(k) implies that P(k+1) will fall), then all dominos will fall.

That we started at 0 is arbitrary; in many cases it's right to start at one. I prefer the term "base case" to "one case" for this reason. It's the first domino that gets knocked down.

I think that makes sense, ty

@Wyman,

so we're given the "one-case" to start our proof by induction...but what do you mean by this: "Let K be the set
{k in N : (m+n) + k = m + (n+k)}. In this proof, we will use two axioms of natural numbers. "

I get your first step, but am confused by the justification for the second step "since k is in K"...

thanks again

I'm saying, let K be the set of all k's for which the statement is true; we'd like to show that K is all naturals.

First we show that 0 is in K. (first domino/base case -- maybe in your case, you show 1 is in K by (A2), whatever)
Now, since K is nonempty (I have no idea how many things are in there, but we showed that 0 was in there), pick an arbitrary guy from that set. Call him k.
Now, we'll show that k+1 must also be in K.

So we've shown 0 in K and (k in K => k+1 in K), so K is all naturals. This is precisely the same statement as
We've shown P(0) and (P(k) => P(k+1)), so P is true for all naturals.

Hm....I think I follow...the domino explanation was very helpful.

So we're trying to figure out for what set this proposition holds. And it turns out that we start out at 0 and show that it holds for each successive case. So the set we that has this property must be all natural numbers...yeah?

yes

So to prove m+n = n+m, where m and n are natural numbers:

Base case: m+1 = 1+ m
m=m

assume m+n = n+m

prove m+(n+1) = (n+1)+m
m+(n+1) = (n+1)+m by the induction hypothesis

is this it?

No. You've assumed what you're trying to prove. This proof actually requires induction on both m and n.

Theorem: For all natural numbers m and n, m+n = n+m.
Proof: We prove this by induction on m. Let
M = {m in the natural numbers : m + n = n + m for all natural numbers n}.
Base case: Notice that 0+n = n = n+0 for all n (this is just the property that 0 is the additive identity; this is somewhere in the axioms). i.e. 0 is in M.
(1st)Inductive Hypothesis: m is in M.
We'll show that m+1 is in M.
To do that, we must show that for all n in the naturals, (m+1) + n = n + (m+1). We'll show this by induction on n.
Let S = {n in naturals: (m+1) + n = n + (m+1)}
Base case: n = 0. We've already seen that 0 commutes with all naturals, so 0 is in S.
(2nd)Inductive hypothesis: n is in S.
We now show that n+1 is also in S.
(m+1) + (n+1)
= ((m+1) + n) + 1, by associativity (which we proved earlier)
= (n + (m+1)) + 1, since n is in S (by our 2nd inductive hypothesis).
= n + ((m+1) + 1), by associativity
= n + (m + (1+1)), by associativity
= n + ((1+1) + m), since m is in M (1st ind. hyp.)
= n + (1 + (1+m)), assoc.
= (n+1) + (1+m), assoc.
= (n+1) + (m+1), since m is in M (1st ind hyp).
Now, since 0 is in S, and since n in S implies (n+1) in S, S is all natural numbers.

Hence we've shown that 0 in M, and m in M implies (m+1) in M, so M is all naturals, proving the theorem.

Notice that we didn't make any "leaps of faith". We only used concrete facts that we knew to be true because we've proved them, assumed them, or have them as axioms.

Hope this helps.

Wow. I think I'm following but isn't the only important thing at the end that M is all natural numbers? I guess I'm confused about why S = all natural numbers is important...

The proof itself makes sense tho, hard to imagine doing that on my own

I'm also a little bit confused about why you're using these sets, M and S...If I'm using induction to show that some series adds up to (n+1)(n+2)/2 or something, I never really use sets like that...any explanation there?

Thanks again brian