Recent content by Applejacks01

If I said A = Integral of e^(-x^2) dx from -infinity to infinity = Integral of e^(-y^2) dy from -infinity to infinity Then A^2 = (Integral of e^(-x^2) dx from -infinity to infinity ) * (Integral of e^(-y^2) dy from -infinity to infinity ) [due to independence of the variables] A^2 = Double...

I know that the indefinite antiderivative doesn't have a closed form solution in terms of elementary functions, but what about definite integrals in polar coordinates? For example, let's take the standard normal. Let's say I want to find P(Z<c), where c is a positive real #. So we have...

Hey guys. I'm sure some of you are aware of how to analytically integrate e^(-x^2) dx from - infinity to infinity using polar coordinates. I have taken that logic and showed that the integral of the normal distribution( not necessarily standard) integrates to 1 over the entire domain...

Okay for example: If 1,2 have the same birthday as 3,4 , then that is considered a quadruple, not 2 pairs. For a group of 4 people, we can have the following cases: (0 same): 1 <> 2 <> 3 <> 4 (1 pair) 1=2 and 3<>4 , 1 = 3 and 2<>4, 1=4 and 2<>3. 2=3 and 1<>4, 2=4 and 1<>3, 3=4 and 1<>2 2 pairs...

Yes! I realized last night I was missing that one! (and now my solution for the 5 person case is correct, thank you) Okay guys, so I thought all was well, but to really test my combinatoric skills I decided to try the 8 person case. Well...once again there is a discrepancy! I will attach my...

Yes. If you let u = 2t, then we'd have sin(u)^2 + cos(u)^2 Which equals 1, and we don't have a "u" to plug back into.

Note the identity sin(kx)^2 + cos(kx)^2 = 1 |v| becomes sqrt(4(1)) = 2

You can derive E(Y^2) easily: let f(x) = ƩC(m,x)*q^x * (1-q)^(m-x) E(X^2) = ƩC(m,x)*x^2 *q^x * (1-q)^(m-x) Observe that this is equivalent to ƩC(m-1,x-1)*(m/x)*q^x*(1-q)^(m-x)*x^2 This becomes: ƩC(m-1,x-1)*(m)*q^x*(1-q)^(m-x)*x Let y = x-1 x = y+1...

Homework Statement What is the probability that given a group of 5 people, at least 2 will share the same birthday? Homework Equations I know that 1-P(0 matches) = answer, but there is a reason I am going about this the head on approach. I am trying to develop my combinatoric skills.The...

Ok assuming y is positive( it actually is for the question I was really working on)... okay so u = x/y implies x = y*u. So we have (y*u*Delta(u-t)) we need to convert dx to du x = y* u Dx = y du int(y*u*Delta(u-t)*ydu) So u = t and we get y*t*y*Heaviside(yt) I can see how the y...

Your sum is incorrect. it should be [1-r^(n+1)]/[1-r] if its from i = 0 to n So subtract the term where i = 0 to get the proper sum from 1 to n

Homework Statement ∫x*delta((x/y)-t) dx from 0 to infinity Homework Equations ∫x*delta((x/y)-t) dx from 0 to infinity = ty*|y|*θ(ty) The Attempt at a Solution Okay, so using the transformation of variables technique via the Jacobian, I see where the |y| comes from. However, using...

Thank you so much Ray, you've been very helpful!

Well actually I think that P(X=2) is Sum C(n,2) *p^2 * (1-p)^(n-2) from n = 1 to 5, and when n equals 1, C(n,2) =0. But to directly answer your question,I would say the values are 0. Sorry for not using latex BTW. I'm on my lunch break. Just trying to learn.

Hi, I actually have one more question. suppose I wanted to calculate the probability that x=#successes =2. Do I need to conditionalize f(n) = distribution of n(# trials) such that n>=2, or do I just assume that the probabilities when n=0 and 1 are 0??