# How to find the volume between y = 2√xz and y = 1?

#### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Find $P(Y^2>4xz)$, where $(x,y,z)\in (0,1)^3$."

2. Relevant equations

3. The attempt at a solution
$P(Y^2>4xz)=P(Y>2\sqrt{xz})+P(Y<-2\sqrt{xz})=P(Y>2\sqrt{xz})=P(1>Y>2\sqrt{xz})$

Here, I set the boundaries for integration over this "wedge", for the lack of a better term.

$2\sqrt{xz}\leq y < 1$
$\frac{1}{4z}\leq x < ...$

I tried drawing a picture, but all I got from the information was that x could not be bounded by 0 or 1. The curve doesn't touch the any of the axes except at the origin.

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#### StoneTemplePython

Gold Member
1. The problem statement, all variables and given/known data
"Find $P(Y^2>4xz)$, where $(x,y,z)\in (0,1)^3$."

2. Relevant equations

3. The attempt at a solution
$P(Y^2>4xz)=P(Y>2\sqrt{xz})+P(Y<-2\sqrt{xz})=P(Y>2\sqrt{xz})=P(1>Y>2\sqrt{xz})$

Here, I set the boundaries for integration over this "wedge", for the lack of a better term.

$2\sqrt{xz}\leq y < 1$
$\frac{1}{4z}\leq x < ...$
well I don't think you stated that $Y$ is uniform taking values in 0,1 but I suspect that it is given that your post title asked for Volume...

Two other things:
1.) The issue in some sense is something seems missing. $Y$ is a random variable and so it presumably comes with a cdf and x, y are simply fixed values, apparently not random variables. There isn't any integration work to be done here if you already have the CDF. Just use the CDF of Y and then negate it and add one to get its complement. That being said, I have a hunch you want X and Z to be uniform random variables as well, though this is not stated anywhere

2.) In terms of pattern recognition this looks to be a touch different, but extremely close to this problem from October

#### Eclair_de_XII

The issue in some sense is something seems missing. $Y$ is a random variable and so it presumably comes with a cdf and x, y are simply fixed values, apparently not random variables.
Yes, they are all random variables distributed uniformly in the unit cube in $\mathbb{R}$.

And yes, that problem is exactly the same, except that the problem I'm doing requires me to find the probability of finding two distinct real roots.

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#### Ray Vickson

Homework Helper
1. The problem statement, all variables and given/known data
"Find $P(Y^2>4xz)$, where $(x,y,z)\in (0,1)^3$."

2. Relevant equations

3. The attempt at a solution
$P(Y^2>4xz)=P(Y>2\sqrt{xz})+P(Y<-2\sqrt{xz})=P(Y>2\sqrt{xz})=P(1>Y>2\sqrt{xz})$

Here, I set the boundaries for integration over this "wedge", for the lack of a better term.

$2\sqrt{xz}\leq y < 1$
$\frac{1}{4z}\leq x < ...$

I tried drawing a picture, but all I got from the information was that x could not be bounded by 0 or 1. The curve doesn't touch the any of the axes except at the origin.

View attachment 233742
Notation matters: I cannot figure out if you mean
$$P(Y^2 > 4 X Z)\hspace{4ex}(1)$$
or
$$P(Y^2 > 4 x z)\hspace{4ex}(2)$$ exactly as written. In (1), $X,Y,Z$ are iid $U(0,1)$ random variables, while in (2) only $Y$ is random.

If you really did mean (1), it is important to remember the distinction between a random variable and a realization of that variable. Random variables are typically denoted by upper-case letters near the end of the alphabet, while realizations of those variables are typically (but, of course, not always) denoted by lower-case versions of the same letters.

If you meant (2) the problem is essentially trivial, with a one-line solution needing no fancy calculations.

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#### Eclair_de_XII

If you meant (2) the problem is essentially trivial, with a one-line solution needing no fancy calculations.
Do you mean that the problem becomes a one-dimensional integration problem if I am not integrating with respect to the random variables $X,Z$?

#### StoneTemplePython

Gold Member
And yes, that problem is exactly the same, except that the problem I'm doing requires me to find the probability of finding two distinct real roots.
I'm not sure whether you finished your last exercise on right continuity of CDFs, but the thought that immediately comes to mind is a direct follow up to it.

conditioning on $4XZ = \alpha$ for $\alpha \in [0,1]$

what is $P(Y^2>\alpha)$ and what is $P(Y^2\geq \alpha)$?
(hint: there are no jump discontinuities in the CDF for the random variable $Y^2$).

The conclusion should be that the exercises are in fact probabilistically the same. I'd formalize this point via conditional expectations with two different indicator random variables, but the basic idea is zero probability of 'collisions' with continuous random variables.

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#### Eclair_de_XII

what is $P(Y^2>\alpha)$ and what is $P(Y^2\geq \alpha)$?
They should be the same, right? No jump discontinuities mean $P(Y^2=\alpha)=0$ and $P(Y^2>\alpha)=P(Y^2\geq \alpha)$.

#### Eclair_de_XII

Anyway, are you asking me to find: $P(Y^2\geq a|\alpha = a)$? I kind of already went ahead and did part of the method described by the person whose topic you linked me to.

$P(X\geq \frac{y^2}{4Z}|Y=y)=\int_{\frac{1}{4}y^2}^1 \int_{\frac{y^2}{4z}}^1 dx dz =\int_{\frac{1}{4}y^2}^1 (1-\frac{y^2}{4z})dz=(z-\frac{y^2}{4}ln(z))|_{\frac{1}{4}y^2}^1= [1]-[\frac{1}{4}y^2-\frac{1}{4}y^2ln[(\frac{1}{2}y)^2]]=\\1-[\frac{1}{4}y^2-\frac{1}{2}y^2ln(\frac{1}{2}y)]$.

And so...

$P(X\leq \frac{y^2}{4Z}|Y=y)=1-P(X\geq \frac{y^2}{4Z}|Y=y)=1-(1-[\frac{1}{4}y^2-\frac{1}{2}y^2ln(\frac{1}{2}y)])=\frac{1}{4}y^2-\frac{1}{2}y^2ln(\frac{1}{2}y)$

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#### StoneTemplePython

Gold Member
Anyway, are you asking me to find: $P(Y^2\geq a|\alpha = a)$? I kind of already went ahead and did part of the method described by the person whose topic you linked me to.
basically, yes. $P(Y^2\geq \alpha\big)$ is merely the complementary CDF of a random variable $Y^2$ (supposing we ignore zero probability events underlying the equality case)

from here you just need the density associated with the random variable $\big(4XZ\big)$ and then to integrate over this density. It's a bit ugly as is common for finding distributions for products of random variables.

#### Eclair_de_XII

I haven't learned how to find the density of the product of two independent r.v.'s, but I did attempt to get a distribution function..

$P(4XZ<\alpha)=\int P(4XZ<\alpha|X=x)P(X=x)dx=\int P(Z<\frac{\alpha}{4x})P(X=x)dx\\ =\int \int_0^{{\alpha}{4x}}dzP(X=x)dx=\int {\frac{\alpha}{4x}}dx=\frac{\alpha}{4}ln(x)$

Even if this were right, I'd still have to determine the bounds of integration w.r.t. $x$.

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#### StoneTemplePython

Gold Member
I haven't learned how to find the density of the product of two independent r.v.'s,
on the computational side, this *is* in some sense, the exercise. Alternative approaches would be to take advantage of positivity and do a log transform of the random variables, sum (i.e. convolve) them, then exponentiate back, or if you want to focus on, in effect, getting the CDF, then what you've done below should ultimately get you there.

$P(4XZ<\alpha)=\int P(4XZ<\alpha|X=x)P(X=x)dx=\int P(Z<\frac{\alpha}{4x})P(X=x)dx\\ =\int \int_0^{{\alpha}{4x}}dzP(X=x)dx=\int {\frac{\alpha}{4x}}dx=\frac{\alpha}{4}ln(x)$

Even if this were right, I'd still have to determine the bounds of integration w.r.t. $x$.
High level this seems like an ok idea. Some cleaup of course is needed e.g. as stated before $P(X=x) =0$-- what you actually want is a density there.

- - - - -
edit:
this may be overkill, but really to get the above, I'd setup an event $B$ that is $4XYZ \lt \alpha$ (note you should be able to confirm that this has the same probability as $4XYZ \leq \alpha$ which is in fact preferable since this is how CDFs are written)

for some arbitrarily chosen $\alpha \in (0,1)$. So what you have is

$P(B) = E\Big[\mathbb I_B \Big] = E\Big[E\big[\mathbb I_B \big \vert X\big]\Big]$

In some sense this is close to what you've done above, but the chaining on of $P(X=x)$ did not quite sit well with me. This approach should make it clear that

$E\big[\mathbb I_B \big \vert X\big]$ is a random variable that takes on values

$P(Z<\frac{\alpha}{4X})$

or if you prefer, for each $X(\omega) = x$ it takes on values
$P(Z<\frac{\alpha}{4x})$

and it should be clear that when you take expectations of this random variable
$E\big[\mathbb I_B \big \vert X\big]$ you are using the density of $X$ -- this is part of a general approach to conditional expectations.

- - - - -

If you want to focus on $P(4XZ<\alpha)$: you could then set this up to have an event $A$ where $4XZ\lt \alpha$ and $\alpha := Y^2(\omega)$

giving you

$P(A) = E\Big[\mathbb I_A \Big] = E\Big[E\big[\mathbb I_A \big \vert Y^2\big]\Big]$

in which case you can get by with the CDF for $(4XZ)$ and then merely integrate over the density of $(Y^2)$

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#### Eclair_de_XII

log transform of the random variables, sum (i.e. convolve) them, then exponentiate back, or if you want to focus on, in effect, getting the CDF
I haven't been taught how to do any of that...

$E[E[\mathbb{I}_B∣X]]$
I'm very confused by this notation every time I see it and have no idea what it is supposed to mean.

#### StoneTemplePython

Gold Member
I haven't been taught how to do any of that...

I'm very confused by this notation every time I see it and have no idea what it is supposed to mean.
You make the job not at all easy for us -- when you leave "relevant equations" blank, we are stuck guessing at what you know and don't. It is the most important part of the template. If you want a belt and suspenders approach, stating books you're using / have used may also help bridge gaps -- this may be overkill, but that's why they call it belt and suspenders.

Since you're in an "advanced probability" course, I simply refuse to believe you've never seen conditional expectations before or that you can't map the notation I gave for the tower property of (iterated) expectations to what you've seen before. There's only a small handful of common notations for this -- I gave one of them. The same for indicator random variables.

edit:
For a reference that uses almost identical notation, consider:
http://www.randomservices.org/random/expect/Conditional.html

in particular (3) and the section called "conditional probability" about a fifth of the way down, though they use $1_A$ for an indicator instead of $\mathbb I_A$ -- it is a very minor difference.

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#### Eclair_de_XII

I'm sorry for all the trouble... I am using Grimmett's Probability: An Introduction, and I'm on chapter six with joint and multivariable distributions, I think it was called. I couldn't think of any relevant equations applicable to the problem. I'm on the section right before the conditional probability.

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#### StoneTemplePython

Gold Member
I'm sorry for all the trouble... I am using Grimmett's Probability: An Introduction, and I'm on chapter six with joint and multivariable distributions, I think it was called. I couldn't think of any relevant equations applicable to the problem. I'm on the section right before the conditional probability.
so it sounds like you will get to conditional probability very soon, and then I can breathe a little easier. The index suggests that conditional expectations come up on page 33 (past) and page 99 (upcoming).

It sounds like they don't emphasize conditional expectations in this text, which is a pity since they are one of the gems of probability theory... (Blitzsteins book does a good job on this and even has a more general variant of your problem, but you don't have control over the book I'd wager.)

As for this problem, I'd suggest returning to
$P(4XZ<\alpha)=\int P(4XZ<\alpha|X=x)P(X=x)dx=\int P(Z<\frac{\alpha}{4x})P(X=x)dx$
I think this is close... one issue is again use density for $f_X(x)$ and the other is you want to, very carefully, convert

$P(Z<\frac{\alpha}{4x})$ into the cdf for Z.

so what you have is
$P(4XZ<\alpha)= P(4XZ\leq \alpha)= \int P(Z\leq \frac{\alpha}{4x})f_X(x)dx$

you want to integrate over X. it is non-negative so you can formally do this over $(0, \infty)$ but you should recognize that $f_X(x)=0$ for $x \gt 1$ which should simplify the calculations. You know the CDF for $Z$ so you should be able to get the desired result fairly easily I think.

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#### Eclair_de_XII

I think this is close... one issue is again use density for $f_X(x)$ and the other is you want to, very carefully, convert

$P(Z<\frac{α}{4x})$ into the cdf for Z.
Okay, so first thing: the marginal density function $f_X(x)$ isn't one? Second thing, this was the closest I could come to a coherent answer: $P(Z>\frac{\alpha}{4x})=\int_{\frac{\alpha}{4x}}^1dz$ and $P(4XZ>\alpha)=\int_{\frac{\alpha}{4}}^1 \int_{\frac{\alpha}{4x}}^1 f_X(x)dzdx$. It's basically the same as before, and I'm afraid I'm not really understanding this problem all that much.

#### StoneTemplePython

Gold Member
Okay, so first thing: the marginal density function $f_X(x)$ isn't one?
for $x \in (0,1)$ it is 1 -- since X is uniform in (0,1). Definitely.

Second thing, this was the closest I could come to a coherent answer: $P(Z>\frac{\alpha}{4x})=\int_{\frac{\alpha}{4x}}^1dz$ and $P(4XZ>\alpha)=\int_{\frac{\alpha}{4}}^1 \int_{\frac{\alpha}{4x}}^1 f_X(x)dzdx$. It's basically the same as before, and I'm afraid I'm not really understanding this problem all that much.
one nit: do you want $P(Z>\frac{\alpha}{4x})$ or $P(Z<\frac{\alpha}{4x})$? setting aside zero probability events they are complementary of course.

Honestly I think you're close. My general view is don't start from scratch. $Z$ is a uniform random variable (in fact all 3 random variables are iid uniforms right?). Can't you just use the CDF for $Z$ here? I'd probably split it the integral into case where $\frac{\alpha}{4x} \geq 1$ (because your CDF always evaluates to one in that case, right?) and otherwise. But I happen to really like CDFs.

The issue is that this problem has some careful reasoning required with events, and is computationally a little bit nasty, though I kind of like the way you've proposed to solve it.

#### Eclair_de_XII

Okay, so:

$P(4XZ>y^2)=\int_{\frac{y^2}{4}}^1 \int_{\frac{y^2}{4x}}^1 f_X(x)dzdx=\int_{\frac{y^2}{4}}^1 (1-{\frac{y^2}{4x}})dx\\ =[x-\frac{y^2}{4}ln(x)]|_{\frac{y^2}{4}}^1=1-[\frac{y^2}{4}-\frac{y^2}{2}ln(\frac{y}{2})]$ and $P(4XZ<y^2)=\frac{y^2}{4}-\frac{y^2}{2}ln(\frac{y}{2})$.

So:

$P(4XZ<Y^2)=\int_0^1P(4XZ<y^2)P(Y=y)dy=\int_0^1 [\frac{y^2}{4}-\frac{y^2}{2}ln(\frac{y}{2})]dy\\ =\frac{1}{4}\int_0^1 y^2(1-2ln(\frac{y}{2}))dy\\ =\frac{1}{4}[\frac{1}{3}y^3(1-2ln(\frac{y}{2}))+\frac{2}{3}\int y^2 dy=[\frac{1}{4}[\frac{1}{3}y^3(1-2ln(\frac{y}{2}))+\frac{2}{9}y^3]_0^1\\ =\frac{1}{4}[\frac{1}{3}(1+2ln(2))+\frac{2}{9}]-\frac{1}{4}(-\frac{2}{3}\lim_{y\rightarrow 0}\frac{ln(y)}{y^{-3}})\\ =\frac{1}{4}[\frac{5}{9}+\frac{2}{3}ln(2)]=\frac{5}{36}+\frac{1}{6}ln(2)$

My notation might be a bit off, but it's mostly because I am not used to working with conditional continuous distributions...

#### StoneTemplePython

Gold Member
$P(4XZ<Y^2)=\int_0^1P(4XZ<y^2)P(Y=y)dy=\int_0^1 [\frac{y^2}{4}-\frac{y^2}{2}ln(\frac{y}{2})]dy\\$
...
two red flags that jump off the page:
i.) $P(Y=y) =0$ because $Y$ (and $Y^2$ for that matter) is a continuous random variable. Again you need a density here.
ii.) you cannot integrate directly over the density of $Y$. You need to either follow change of variables formulas or compute the density of $Y^2$ and use that. It's very easy to mess this up -- and this is part of the reason why I fall back on the conditional expectations formalism -- it helps me avoid this pitfall.

My notation might be a bit off, but it's mostly because I am not used to conditional continuous distributions...
Honestly this is a pretty brutal problem to insert *before* the conditional probabilities section.

#### Eclair_de_XII

You need to either follow change of variables formulas or compute the density of $Y^2$ and use that.
So I'm going to have to compute: $P(4XZ<Y^2)=\int_0^1 P(4XZ<y^2)f_{Y^2}(y^2)dy$?

"How to find the volume between y = 2√xz and y = 1?"

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