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1d
comment How to find minimal polynomial of primitive element (field theory)
@brvh If $\alpha$ is an element of order $n$ (note: not just an $n$-th root of unity) by which we mean that $n$ is the smallest positive integer such that $\alpha^n = 1$, then $\alpha^k$ is an element of order $\frac{n}{\gcd(n,k)}$. If you do not know this result, try to derive it for yourself and remember it for future use; it is very handy.
1d
revised How to find minimal polynomial of primitive element (field theory)
added 233 characters in body
1d
answered How to find minimal polynomial of primitive element (field theory)
2d
comment Generator polynomial creates a 127 bit sequence
The terms you need to search for are linear feedback shift register, maximal-length sequence, PN sequence, pseudonoise sequence, linear recurrence etc.
May
21
comment Density function of minimum of random variables
Hint: $P\{\min_i X_i > a\} = \prod_{i=1}^n P\{X_i > a\}$ and so you can get the expected value directly without explicitly finding the density of the minimum. @Frank
May
20
comment Prove absolute sum expectation
Hints: $\sum_i X_i$ has $0$ mean and variance $n$. The CLT says that the CDF of $\frac{1}{\sqrt{n}}\sum_{i=1}^n X_i$ is converging to the CDF of a $N(0,1)$ random variable as $n\to\infty$. For a nonnegative random variable $Y$ such as $\left|\frac{1}{\sqrt{n}}\sum_{i=1}^n X_i\right|$, $\displaystyle E[Y] = \int_0^\infty [1-F_Y(y)]\,\mathrm dy$
May
20
comment Proof conditional probability formula
Yes!!! Now, please copy your calculation above into an Answer (yes, you are not just permitted but actually encouraged to post an Answer to your own Question, and even Accept it as the best answer if you wish). That way, this Question will not remain in the Unanswered category and be thrown up repeatedly as one of the top questions of the day.
May
19
comment Proof conditional probability formula
You are nearly there. Combine the two expressions that you have so diligently worked out to state what ratio of probabilities $p_{X\mid Y, U}(i \mid j,k)p_{U\mid Y}(k \mid j)$ works out to be. If you remember that $\Pr(C \cap D) = \Pr(D \cap C)$, you may be able to cancel some terms. If you do get something of the form $\frac{\Pr(E \cap F \cap G)}{\Pr(H)}$, see if you can sum the resulting expression from $k = 0$ to $\infty$. (Hint: $\Pr(H)$ does not depend on $k$ at all and so can be pulled out as a constant so that you are just summing $\Pr(E \cap F \cap G)$ from $k=0$ to $\infty$.)
May
19
comment Proof for Mean of Geometric Distribution
@Starlight Oh, cut the crap! The top item in the banner (the one immediately above the one you give above) says "This article needs attention from an expert on the subject. The specific problem is: Mathematics." To quote your words back to you; could you be a little more factual?
May
19
comment Proof conditional probability formula
Better and better. So now you have that $$p_{U\mid Y}(k\mid j) = \Pr(U=k \mid Y = j) = \frac{\Pr(\{U=k\}\cap\{Y=j\})}{\Pr(\{Y=j\})}.$$ Can you write down a similar expression for $p_{X\mid Y, U}(i\mid j, k)$ without additional help and prompting? Hint: the numerator will be the probability of the intersection of three events,_ the denominator the probability of the intersection of two events.
May
19
comment Proof for Mean of Geometric Distribution
@Glen_b The OP has not actually found a fourth way of summing the series (other than by blindly applying a formula that he found on a Wikipedia page which itself admits, in a banner at the very top, that there are some unresolved mathematical issues in the contents of that page).
May
19
comment Proof for Mean of Geometric Distribution
@Fato39 But perhaps the fact that there is a banner at the top of the Wikipedia page in question stating that there are some unresolved mathematical issues that still need to be considered might give the OP pause...
May
18
comment Proof conditional probability formula
Better! Now, $\Pr(U=k\mid Y = j)$ is a conditional probability, and as such is defined to be the ratio of two (unconditonal) probabilities $P(A)$ and $P(B)$. That is, $$Pr(U=k\mid Y = j) = \frac{P(A)}{P(B)}.$$ What are the events $A$ and $B$?
May
18
comment Proof conditional probability formula
Good! Now, let's try and figure out what events $A$ and $B$ are meant when we use $p_{U\mid Y}(k\mid j)$ as an expression for $P(A\mid B)$.
May
17
comment Proof conditional probability formula
Do you have, in your notes, or in your textbook, or even on the problem set you are working on, a definition of probability mass function? Not the name "probability mass function", or the abbreviation "pmf", or the notation "$p_X(a)$", but what it all means? If your definition says that $p_X(a)$ equals something, tell us what it equals. If your definition says "The probability mass function (pmf) of a discrete random variable is $p_X(a)$." and stops there without further elucidation, then I do not think I will be able to help you further.
May
17
comment Proof conditional probability formula
No, don't tell me about functions, which is a word whose meaning I do not understand. $p_{X\mid Y, U}(\mid j, k)$ is a number, say $0.0347$, that is a probability of some kind. What kind? and what event is $0.0347$ the probability of? Your response needs to be something like $P(A) = 0.0347$ or $P(A\mid B) = 0.0347$ where you need to replace $A$ and $B$ by something that might be stated in terms of $X, Y, U$ and possibly $i, j, k$.
May
17
comment Simple Question on Conditional Probability with Two Events
Welcome to math.SE. You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
May
17
comment Proof conditional probability formula
$p_{X\mid Y, U}(i\mid j, k)$ is a probability of some kind. So, what is it the probability of?
May
16
answered Let $X$ ~ $Exp(1)$ and $Y$ ~ $Exp(2)$ independent random variables. Let $Z = max(X, Y)$. Calculate expected value of $Z$.
May
16
comment problem about graph of auto-correlation for wide-sense stationary process?
This question from your exam is based on nonsensical premises. The autocorrelation function of a wide-sense-stationary process cannot have the shape shown in the figure: there should be a peak at the origin (or the process is trivial with $X(t) = Y$ for all $t$.