Recent content by Rashad9607

Original homework friend had a nice comment: [FONT=arial]I think one aspect of the "naturalness" of the "point + radius" parametrization is that it is continuous - changing any of the parameters transforms shapes smoothly.

You could reconstruct the right order of digits of the original numbers, but you would lose positive/negative and decimal point placement. Also you have to define the range of the function before you can qualify it as a bijection. I did skip over decimals and negatives because it adds tedium...

Agreed- also there are triples of points giving rise to no circle. In the three point/six parameter system, given two points, there are constraints on what the third point can be. We might reason that taking advantage of whatever structure exists may allow us to more efficiently encode a...

A friend's homework problem (Prove any five points in the plane determines a possibly degenerate conic section) led us to a different problem that we found more interesting. We can identify a circle with three points on the circle, or six parameters $(x_1,y_1,x_2,y_2,x_3,y_3)$ where, keeping...

So I have read the next section in my text and learned that a characteristic of exponential family distributions is that the values $x$ can take must be the same over the entire parameter space. If we take $\theta=(p,n)$, then x=0,1,2,...,n, which depends on $\theta$, so it cannot be an...

Re: Dicrete-continuous random variable! Cool problem. Assign $Z_{i}$ to have mean $\mu_{i}$ and pdf $f_{i}(x)$. Everything should follow if we find a pdf $f(x)$ for $X$. $f(x)=P(X=x)=P(X=x|Z_{1})P(Z_{1})+P(X=x|Z_{2})P(Z_{2})$ (Law of Total Probability) $=pf_{1}(x)+(1-p)f_{2}(x).$...

A pdf is of the exponential family if it can be written $ f(x|\theta)=h(x)c(\theta)exp(\sum_{i=1}^{k}{w_{i}(\theta)t_{i}(x))}$ with $\theta$ a finite parameter vector, $c(\theta)>0$, all functions are over the reals, and only $h(x)$ is possibly constant. I would like to show the binomial...

The goal for me is not to verify the problem as many times as possible but to learn some probability from it. Mutual independence is a completely different argument, a stronger result, and, I think, a more direct method, but I don't know if the argument works. Therefore I have something to...

Thanks, ILikeSerena. Here's another attempt... Take any subset $\Lambda$ from $\{A_{1}^{C}, A_{2}^{C}, A_{3}^{C}, ...\}$. We can label this subset $\Lambda=\{A_{j_{1}}^{C}, A_{j_{2}}^{C}, A_{j_{3}}^{C}...\}.$ Consider $P(\bigcap_{\lambda\in\Lambda}\lambda)$...

Does this look okay? $P(\cap{A_{i}})<1\Rightarrow 1-P(\cap{A_{i}})=P((\cap{A_i})^{C})=P(\cup{A_{i}^{C}})>0.$ $P(\cup{A_{i}^{C}})>0\Rightarrow\exists j\in{\mathbb{N}}$ with $P(A_j^{C})>0$. (Consider that $P(\cup{A_{i}^{C}})\leq\sum{P(A_{i}^{C})}$) Then $P(A_{j})=1-P(A_{j}^{C})<1$...

Hello all, I don't recall where we originally found the problem but we certainly had some fun with it before I offered it to Jameson. A similar problem was posted on stackoverflow, of all places. Here was my lamentable first solution: $f(x)=\left\{\begin{matrix} 0 & | &x=0 \\...

In this case it is simple to see whether the line segment intersects the circle; compare the line connecting the origin to (12,16) and the line connecting the origin to the center of the circle.

Heed Avodyne's advice! Say you were asked to reduce 64/8 and you decided to multiply by 8 to get rid of the denominator. You now have a different number. Don't confuse this with equations, where you can perform an operation to both sides (like division) and maintain the equality.

Maybe it well help to give you a similar problem. Say you have ten pens and two are broken. How many sets of one member are there of broken pens? Well, two, because there are two broken pens. How many sets of one member are there of functioning pens? Eight. Now...how many sets of two are there...