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2d
comment Joint distributions where one is uniform
You wrote "the bounds that $F_Y(y)$ would take." That's not what you meant, but the values of $F_Y(y)$ range over $[0,1)$.
2d
comment Joint distributions where one is uniform
$X$ is uniform on the open interval $(0,1)$. When $X$ is close to $0$, $X/(1-X)$ is close to $0$. Note that it is "never" $0$. When $X$ is close to $1$, $X/(1-X)$ is large positive. So the range of values for which the cdf of $Y$ is given by the formula of the answer is $(0,\infty)$. Almost what you wrote, except we don't include $0$. The cdf is $0$ for $y\le 0$.
2d
revised Joint distributions where one is uniform
added 1 character in body
2d
answered Joint distributions where one is uniform
2d
answered If $S$ and $T$ are subrings of $R$, is $S+T$ a subring of $R$?
2d
comment Relationship between operations of a ring
The integers modulo $m$ come close.
2d
comment Size of a point.
Points have no part, in (very) old-fashioned language, roughly speaking means that points have no non-trivial subset. (Here I am being very anachronistic.) Points have no part, but they can certainly be part of larger assemblages, such as lines.
2d
revised Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
edited body
2d
answered Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
2d
comment Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
But at least the finding the polynomial is finished.
2d
comment Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
It is easiest with some theory, using the fact that your polynomial is irreducible over the rationals. If we don't have theory, it is s mattter of proving your number is not the root of a quadratic with rational coefficients. Kind of like the proof of irrationality of $\sqrt{2}$ but a lot messier. If it is a graded mechanically contest problem (full credit for correct answer, 0 otherwise) you need not worry about it.
2d
awarded  Enlightened
2d
comment Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
The first attempt (cubing directly) came a lot closer. We have $x^3=56+3(7^{5/3}+7^{4/3})=56+21x$. Now deal with the $4$ business, and you will have your polynomial. Then (depending on the source of the problem) you may or may not have to prove minimality.
2d
comment Urn with white and black balls, probability of n white with nth position
The second problem can be solved using basically the same method as the first problem.
2d
comment picking balls from $3$ jugs
The procedure is right. I have not checked the arithmetic. For $4$ black you could have used the symmetry.
2d
comment Computing the GCD of two polynomials
Are you working in $\mathbb{R}[x]$ (or $\mathbb{Q}[x]$) or $\mathbb{Z}[x]$? If you are working with polynomials over a field, there are technically infinitely many gcd, for we can multiply by any unit. However, the convention is to use a monic polynomial.
2d
answered How to calculate this derivative? Do I have to use the chain rule?
2d
comment Difficulty in finding marginal distribution
By symmetry the conditional expectation of $X_1$ given $X_2$ is $0$. The random variables are not independent. The joint density, when you take account of the part where it is $0$, is not a product. Joint density a product only works when the region is a rectangle (possibly infinite) with sides parallel to the axes. Much more informally, if you know that $X_2$ is "small" then you know that $|X_1$ cannot be large, so information about $X_2$ gives you some information about $X_1$.
2d
comment If $(d,a)=1$ and $d|ab$ then $d|b$ .
The proof of the Euclid Lemma in the prime $p$ case needs a little more than for your current result, for we need to show, from $p$ does not divide $a$, that $(a,p)=1$. Easy, but requires a sentence. But since in the current problem we are given that $(d,a)=1$, that part of the proof of Euclid's Lemma is a) unnecessary and b) if you put it in, it will look as if you don't know what's going on.
2d
awarded  Nice Answer