Second investor is more risk averse.

Basic Real Analysis Question

I'm a bit confused on the difference between image and preimage.

It seems like the image is the set of all of the elements in the range that get mapped to. And the preimage is the set of all of the elements in the domain that map to elements in the range.

Usually, we're asked what's the image of [-2, 2] given some function or the preimage of [1, 4] given some function.

Do I have it right?

Yes. It's not necessarily real analysis, it's basic to any function.

ty, lastcard

Right, think of a function f:A-->B as a bipartite graph (treat A and B as disjoint) with edge (a,b) whenever f(a)=b. Then image and preimage are just neighbourhood of a given set of vertices.

Im working on an eigenvectors question and ive been given a 3x3 matrix.

Ive got down to -x^3 + 8x^2 -3x; my mind has gone blank and i cant factorise it. Someone throw me a bone?

One of the factors is x.

This is probably a very easy question for anyone halfway decent with math, but appreciate any help.

If I have a number. Let's say $350 is 7% of another number. How do I find the original number?

Thanks.

Let x be the original number. Then 350 = (7/100)x so x = 100(350)/7.

^^^Thank ya brutha!

-x^3 + 8x^2 -3x = 0

-x*(x^2 - 8x + 3) = 0

x = 0 or x = 4 +/- sqrt(13)

Why?

Investor A
Return 10% stdev 15%
Return 18% stdev 20%

Investor B
Return 12% stdev 15%
Return 15% stdev 20%


Investor A appears to have a steeper Return/stdev curve.


PairTheBoard

ya im bad at arithmatic or i just misread it at the time.

OK probability theoryish question for you.

If you have a joint PDF of 2 random variables X and Y.

From my understanding in order to determine if X and Y are independent then the product of their marginal PDFs MUST be equal to their joint PDF.

This is clear, however I also see that another way to determine independence of a joint PDF is if you can break the joint PDF into a product of any two functions. IE.

f(x,y) = g(x)h(y) <-- if this can be determined then we also have independence.

but finding any two functions g(x)h(y) doesn't imply that g(x) and h(y) are necessarily X and Ys marginal PDF?

Are they both ways to find independence of random variables and if so is their always a relationship between marginal pdfs any two functions g(x)h(y) we could derive?

hey can someone help me prove that cosx/tanx= (1/sinx)-sinx, by expressing the left side in terms of sinx. Thanks

cosx/tanx = cosx/(sinx/cosx) = cos^2x/sinx = (1-sin^2x)/sinx, etc.

sorry im not very good at this and forgot a lot of it, where should i got from there?

No problem, I like doing math.

(1-sin^2x)/sinx = 1/sinx - sin^2x/sinx = 1/sinx - sinx

Thank you very much sir. Don't get me wrong i really enjoy math, just right now i'm not very good at it....

How do you know whether or not an equation is a function by just looking at the equation. For example, how do you know y^2= 4-x^2 is not a function. I know what the definition of a function is, I just don't understand how you figure this out when looking at an equation. Thanks in advance.

An equation is not a function.

The solution set of such an equation can define a function f(x) if (x0,y1), (x0,y2) in the solution set implies y1=y2. ie. One value of x can only yield one possible value of y solving the equation. That unique value of y is how f(x) is defined. The values of x for which solutions exist are then the domain of the function so defined.

Notice that'a not the case here. eg. For x=sqrt(2) there are two possible values of y which solve the equation +sqrt(2) and -sqrt(2).

PairTheBoard

Intuitively, the idea is that a function f:X->Y maps each element of X to exactly one element of Y. More fundamentally, a function f:X-->Y is a set of ordered pairs (x,y), with x∈X and y∈Y, where (x,y)∈f means that f(x) = y. To satisfy the condition that each of point of X is mapped to exactly one element of Y, we impose the condition that if (x,y)∈f and (x,z)∈f, then it must be the case that y=z. (Otherwise, we would have f(x) = y and f(x) = z, where x would be getting mapped to two different points under f.)

Similarly, the solution set to an equation such as y^2 = 4-x^2 will be a set of ordered pairs. For example, the solution set to the equation you gave is {(x,y)∈ℝ^2 | y^2 = 4-x^2}. So, to determine whether or not the solution set to an equation defines a function, you have to determine whether or not for every (x,y) pair satisfying the equation, there is no other number y_2 such that (x,y_2) satisfies the equation.

You may have been taught a heuristic method for figuring this out that was called something like "the vertical line test", where you would graph the solution set to an equation and consider drawing vertical lines through it. If you could draw a vertical line that intersected the graph at more than one point, then the graph could not define a function. Going back to the earlier explanation, this is because you would be mapping an "x" to more than one "y".

For example, consider your equation: y^2 = 4-x^2
Rewriting this, we have x^2+y^2 = 4. This is simply the equation of a circle of radius 2 centered at the origin. Using the vertical line test, we see that each line intersects the graph of the solution set in two spots. This is because the square root function is multi-valued, as PTB points out. In particular, we observe that x=0, y=2 satisfies the equations, but so does x=0, y=-2. Consequently, the solution set of this equation cannot define a function.

Question on matrices

Determine a left inverse of the matrix

1 1 1 1] T
-1 1 2 2
[1 2 1 3

Okay, so im unsure on whether to leave the matrix as the transpose (A^T) or find A and solve.

Once we have our matrix A, we let L =

[ 1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1] ?

Then do row operations to get A into triangular form, then LA should = the identity matrix?

Hey just 2 quick trigonometric identity problems confusing me if anyone has time:
prove that tan^2x - sin^2x = (sin^2x)(tan^2x)
also prove that (sinx + cosx)(tan^2x+1/tanx) = 1/cosx + 1/sinx

Thanks.