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Random graphsGiveI positive integer Llc probability 0 < p < 1 (which could function o n), Lfle T:UId(In graph G(n,p) is graph G = (V,E) o |VI = verGices which...

Question

Random graphsGiveI positive integer Llc probability 0 < p < 1 (which could function o n), Lfle T:UId(In graph G(n,p) is graph G = (V,E) o |VI = verGices which every possible edge is present with probability p adl these €VCItS are indcpendent . The edgc probability could bc COIISGAlIt independent of n (sy could also depend On n like p(n) logln) 9) , but (The sunple space is thc sel of all graphs OL n verlices and Lhe probability of aHLY graph depends On its number of edges_ In Lhis projec

Random graphs GiveI positive integer Llc probability 0 < p < 1 (which could function o n), Lfle T:UId(In graph G(n,p) is graph G = (V,E) o |VI = verGices which every possible edge is present with probability p adl these €VCItS are indcpendent . The edgc probability could bc COIISGAlIt independent of n (sy could also depend On n like p(n) logln) 9) , but (The sunple space is thc sel of all graphs OL n verlices and Lhe probability of aHLY graph depends On its number of edges_ In Lhis projecL You investigate VATIOIS properties (f random graphs_ Yor results should include: (a) Show that for every < 1/1000 aud p = cn; with probability that Lends to tends Lo infinity; all the cOnnected €OIponents o G(n, pln)) arc Of size O(log n). (this is actually true for all c < 1, CAIL YOIL prove it?) (b) Show that for eVery 2 1000 and p = c/n; with probability Ghal tends to aS TL tcnds to infinity; Lhere exisbs connecbed COponenl of a linear SIZC. (this is true for all c > 1, CL VOu proC something likc that?) For which p (aS fiicLion p(n) of n) , does the raudom graph G(n, P(n) ) have no isolated vertex (i.e; vertex with no edge incident to it)? Thc type of result yOn should prove is that if p(n) > (1|c)log" Lhen Lhe probability Lhat ((n; p(n)) has a1 isolated vertex tcnds to 0 aS #l Lends G0 infinity (say for fixed while if p(n) log " thcn the probability that G(n, P(n) ) has isolated verlex Lends to 13 It is useful Lo look at the randm valriahle representing Ghc numnbcr of isolated verlices and Lo compute its expectation; variance, ec LolI Show GhaL for p > (1 | c) LheI Ghc probability Lhat G(n, P) is connected tends to #S m Gcncs Lo infinity Writc (ad prove) result stating thc xize: of thc largest clique in the randoI graph G(n; 4). Again Lhis nceds to be formalized. For which valucs of p do we start scing triangles with probability Lhat tends to 1? Extensions YOI (Ould investigate would be when (i.c for which vallues of p) raIldom graph hals at least OHC triangle clique of size 3), OI Wfcn raLlIdom graph is cOected; or Lhe IaLXim (OI IniniIun) degree of raLIIdo graph: There arc countless other questions thal could be considered_ Peoplc: make creeTs Out of this.



Answers

Let $X(t)$ be a WSS random process.
(a) Show that for any constants $a_{1}, \ldots, a_{n}$ and any time points $t_{1}, \ldots, t_{n}$
$\operatorname{Var}\left(a_{1} X\left(t_{1}\right)+\cdots+a_{n} X\left(t_{n}\right)\right)=\sum_{j=1}^{n} \sum_{k=1}^{n} a_{j} a_{k} C_{X X}\left(t_{j}-t_{k}\right)$
(b) Explain why a valid autocovariance function must be positive semi-definite, i.e., $C_{X X}(\tau)$
must satisfy $\sum_{j=1}^{n} \sum_{k=1}^{n} a_{j} a_{k} C_{X X}\left(t_{j}-t_{k}\right) \geq 0$ for all constants $a_{1}, \ldots, a_{n}$ and times $t_{1}, \ldots, t_{n}$
(c) Now consider the "rectangular function" defined by
$\operatorname{rect}(\tau)=\left\{\begin{array}{ll}{1} & {|\tau| \leq 1 / 2} \\ {0} & {\text { otherwise }}\end{array}\right.$
Show that rect(\tau) satisfies the properties of an autocovariance function expressed in the main
proposition of this section.
(d) By considering $n=3, a_{1}=a_{3}=1, a_{2}=-1, t_{1}=-3, t_{2}=0, t_{3}=.3,$ show that rect $(\tau)$
is not positive semi-definite and, hence, cannot be the autocovariance function of any WSS
random process.
[Note: It can be shown that positive semi-definiteness completely characterizes the collection of all
valid autocovariance functions. That is, any valid autocovariance function is automatically positive
semi-definite, as shown in part (b), and positive semi-definite function $C$ there exists a WSS
random process whose autocovariance function is $C(\tau) . ]$

This example explores rectangular distribution, also known as continuous uniform distribution. And after a lengthy explanation, were given a couple of formulas where basically we have a rectangle and a rectangle is on the closed interval alpha beta, and then we have another interval inside there of A and B. And we want to calculate the probability that something chosen at random, some value chosen at random from the interval of A to B will actually be in alpha beta. So this particular problem, alpha and beta already set for us 0.150 point 065 and then A and B are determined by the problem itself. So in the first example, were asked, what's the probability of getting a pellet that is Greater than or equal to 0.50 mm. So that means a being a smaller value will be 0.0 50 and be can't be any bigger than beta. Now I need to use my formula B minus A over beta minus alpha. And when I evaluate those, I end up with about 0.30. Now, in the next situation, I'm looking for the probability that X is less than or equal to, I need zero point 040 So in this case A will be the smallest value possible on my interval which is alpha and be will be the value in question. And then plugging the information in to my formula and evaluating that expression, I get 0.50 and then and see, I actually want to know the probability that X will be between these two measurements. So now A and B are both different from alpha and beta. And it's just a case again of substituting the values in the right place. In my formula B minus A divided by peter minus alpha, calculating that because these are equal signs. And then the next question I'm asked to find the mean, or mu. So I'm going to use this formula alpha plus beta divided by two. So the mean was Alpha Plus Beta Divided by two. So in our case zero 015 plus zero point 065 divided by two. And then the standard deviation. And that formula was beta minus alpha, divided by The Square Root of 12, Divided by Radical 12. And that is approximately .0144.

This example explores rectangular distribution, also known as continuous uniform distribution. And after a lengthy explanation, were given a couple of formulas where basically we have a rectangle and a rectangle is on the closed interval alpha beta, and then we have another interval inside there of A and B. And we want to calculate the probability that something chosen at random, some value chosen at random from the interval of A to B will actually be in alpha beta. So this particular problem, alpha and beta already set for us 0.150 point 065 and then A and B are determined by the problem itself. So in the first example, were asked, what's the probability of getting a pellet that is Greater than or equal to 0.50 mm. So that means a being a smaller value will be 0.0 50 and be can't be any bigger than beta. Now I need to use my formula B minus A over beta minus alpha. And when I evaluate those, I end up with about 0.30. Now, in the next situation, I'm looking for the probability that X is less than or equal to, I need zero point 040 So in this case A will be the smallest value possible on my interval which is alpha and be will be the value in question. And then plugging the information in to my formula and evaluating that expression, I get 0.50 and then and see, I actually want to know the probability that X will be between these two measurements. So now A and B are both different from alpha and beta. And it's just a case again of substituting the values in the right place. In my formula B minus A divided by peter minus alpha, calculating that because these are equal signs. And then the next question I'm asked to find the mean, or mu. So I'm going to use this formula alpha plus beta divided by two. So the mean was Alpha Plus Beta Divided by two. So in our case zero 015 plus zero point 065 divided by two. And then the standard deviation. And that formula was beta minus alpha, divided by The Square Root of 12, Divided by Radical 12. And that is approximately .0144.


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