Math Lover

Yes that is correct.

yes $f_y(y)$ is a piece-wise function $f_1(y)$ for $-16 \lt y \le 0$ and $f_2(y)$ for $0 \lt y \lt 26$. You need to fix the joint density though.

please see the comment that says the question is from an ongoing contest - math.stackexchange.com/questions/4393012/…

Can some of you look at this question? I felt there was not enough context and very little effort to solve. Not sure if you see the same way... I see it has already got 5 upvotes. math.stackexchange.com/questions/4373800/…

3 answers and 2 deleted answers to a PSQ math.stackexchange.com/questions/4371394/…

btw how do you conclude in your answer that OP + OP' < PQ?

@Anonymous +1 for your answer. btw an upper bound that shows OP + OQ + OR < 3 is sufficient to prove there are no integer solutions.

@petaarantes I have edited the answer for now. Please check. Will wait for Anonymous to respond / post an answer.

you are absolutely right and that is a very good catch

@petaarantes please see comments from Anonymous. There is a stronger lower bound on the value of $OP + OQ + OR$ than what I have in my answer. So it seems there is no integer value possible for $OP + OQ + OR$. Can you please unaccept my answer? I would like to delete it.

@Anonymous you may want to add an answer showing no integer value is possible.

$OP + OQ \gt PQ$ yes and $PQ = AB/2$ yes but how does that lead to $OP + OQ + OR \gt 2$? All we know is $OP + OQ + OR \gt OR + AB/2$

@Anonymous not sure I follow you. Can you please elaborate?

@petaarantes you are welcome!

oh ok, just noticed :)

@user ok, pls see Section Inradius, exradii, and circumradius in en.wikipedia.org/wiki/List_of_triangle_inequalities and link#35 at the bottom

For $s \gt 2R + r$, please see Circumradius, Inradius section in en.wikipedia.org/wiki/Acute_and_obtuse_triangles

@MaryStar any more questions on this? Also, if my answer helped, can you please consider accepting it? :)

@MaryStar yes because t is in [-1. 1] and z-t is in [0, 5] so z is in [-1, 6]

@MaryStar int_{-1}^z (1 / 2) \cdot (1/5) dt = (z - (-1)) / 10

finally the answer should be f(z) = (z + 1) / 10 for -1 < z < 1; f(z) = 1 / 5 for 1 < z < 4 and f(z) = (6-z) / 10 for 4 < z < 6

@MaryStar -1 < z < 1 and 1 < z < 4 are correct. But for 4 < z < 6, it should be \frac{1}{2}\cdot \int_{z-5}^1 f_{X_2}(z-t) dt

yes but when we write pdf, we have to write in intervals of z as pdf is different in different intervals

yes consider the intervals of z and write bounds of x accordingly. I just wrote one for -1 \lt z \lt 1 above

@MaryStar for example, for z in (-1, 1), f_Z(z) = \int_{-1}^z 1/2 * f_Y(z-t) dt and f_Y(z-t) = 1/5

It is correct but bounds of x is not correct. it is different bounds for different values of z, which is what is in my answer.

@MaryStar no that's not fully correct. As I explained with the diagram, pdf is different for different sub-domains of Z. Where are you taking care of that?

That's why we call it convolution - it is operation on two functions that gives a third function. Essentially F_Z(z) = P(Z < z) = P(g(X, Y) < z) and here g(X, Y) = X + Y. Now we write this as P(X + Y < z) = P(Y < z - x)...

How do you get that formula is a different question. For this question, it suffices that it works. See here - en.wikipedia.org/wiki/Convolution_of_probability_distributions or see here, just after example 5.27 probabilitycourse.com/chapter5/5_2_4_functions.php

You don't have to write double integral. Knowing $F_Y(z-x) = \frac{z-x}{5}$, you could write as $ \displaystyle \int_{-1}^z \frac{z-x}{5} \cdot \frac 12 ~ dx$

because $ \displaystyle F_Z(z) = \int_{-\infty}^{\infty} F_Y(z-x) f_X(x) ~ dx$, $f_X(x) = \frac 12$ and $ \displaystyle F_Y(z-x) = \int_0^{z-x} f_Y(y) ~ dy$

If $-1 \lt z \lt 1, F_Z(z) = P(Y \lt z - x) = $ $ \displaystyle \int_{-1}^{z} \int_0^{z-x} f(x, y) \, dy \, dx$. Note $f(x, y) = \dfrac{1}{10}$

on your last question - (cos theta, - sin theta, 1) works right?

I need to go now... if there is more we can discuss later... I am really late and need to wake up in a few hours. I hope someone answers your question for arbitrary manifolds

yes

now for your question on x + y + z = 1 if I assume normal pointing in positive z direction then the boundary will go from (1, 0, 0) to (0, 1, 0) to (0, 0, 1) and to (1, 0, 0)

yes and also to mention, we typically seek orientation given in the question and then accordingly decide orientation of the boundary curve... many a times questions do not mention then we assume outward normal vector and then accordingly orientation of the boundary curve.

yes it is correct

ok we are getting somewhere now

it cannot point horizontally as that would be a cylinder then

yes but that is why we call it generally downward...

take a cup that is opening up.. that is radius of cross section increases as you go up. place a coaster as tangent plane at a point and now take a line perpendicular to the plane at the point of tangency. Now there are two directions... one that takes you inside the surface and one that takes you away from it. the one that takes you away is pointing in which direction? down or up?

when you point the thumb downward, your fingers curl in clockwise direction

below z = 0, the outward normal (take a vector from origin to a point on the sphere below z = 0) is pointing generally downward (negative z direction) so it should be clockwise

cross sections

yes but that does not matter. do you agree that as z increases above 0, the radius of the sphere decreases?

paraboloid is opening up... sphere above z = 0 is not opening up... it is rather folding and at z = 1, x^2 + y^2 = 0

I think it was z = 1/sqrt2.. but that doesnt matter

for the boundary at z = 1/2, you know the outward normal vector is pointing generally upward so curve has to be oriented counter-clockwise

think a tangent plane to this surface at any point and a vector perpendicular to the tangent plane pointing away from the paraboloid surface