Joint and Conditional PDFs in 3 variables

[orchid wind]

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[pale temple]

Desmos 3D

Also, a funny thing about your problem is that I think it is solvable even without knowing the marginals of Y and Z

Or joints

But I am not quite sure, so I will withhold this for now

[pale temple]

For f_Z

It's tricky but I think f_Z should be equal to f_Y

So you are integrating wrong

You integrate for x from z to 1

But in reality the inequality would be that z <= 1 - x

i.e. x <= 1 - z

So in fact you should be integrating from 0 to 1 - z

Also it's odd that you are computing the marginals

It seems to be of virtually no use

[pale temple]

Ah ok

I guess that makes sense

Well

To be fair, if the goal was just to find E[Y] and E[Z] and E[YZ], there is no need to go through all that trouble

Since, for instance, E[Y] = E[E[Y | X]] =E[X/2] = (1/2)/2 = 1/4

similar computation for the other ones

This one, let me think

I will try to get the marginal of Z

since you did the one with Y correctly

It seems that you have obtained the joint of X and Z being:

$f_{XZ}(x, z) = 1_{{z \leq 1 - x}} \frac{1}{1-x} = 1_{{x \leq 1 - z}} \frac{1}{1-x}$

[digital valve]

Rion

[pale temple]

So as you integrate on x:

$f_Z(z) = \int f_{XZ}(x,z)dx = \int_{0}^{1-z} \frac{dx}{1-x}$

[digital valve]

Rion

[pale temple]

Now, you solve the integral, which yields:
$$f_Z(z) = \left[ - \ln(1-x)\right]_{0}^{1-z} = - \ln(z) + \ln(1) = -\ln(z)$$

[digital valve]

Rion

[pale temple]

That's more like it

So that confirms that Z and Y have the same distribution, which was stated in the beginning

For the probability of P(Y+Z <= 1/3)

I think it might be in your favor to use conditional expectations as well

$P\left(Y+Z \leq \frac{1}{3}\right) = E\left[ 1_{{Y+Z \leq 1/3}} \right] = E\left[ E\left[ 1_{{Y+Z \leq 1/3}} \vert X \right] \right]$

[digital valve]

Rion

[pale temple]

This helps a lot because now Y and Z are independent conditionally to X

It's an indicator function

1_{condition} is equal to 1 when the condition is fulfilled

0 otherwise

Sure you're welcome

Since you aren't all too familiar with the notations, I also want to clarify something

$E\left[ 1_{{Y+Z \leq 1/3}} \vert X \right]$ is a random variable

[digital valve]

Rion

[pale temple]

it's a function of X

which you can take the expectation of

but it's possible to compute what this random variable is exactly in our case

because we know fY|X and fZ|X, and also know that Y|X and Z|X are independent

[austere canopy]

Joint and Conditional PDFs in 3 variables

[pale temple]

Page 2, I don't know why you change the bounds of the integral

There is no reason for you to do so

I mean I understand the motivations for you to do so

But the writing is not good

The pdf of (X,Y,Z) should be:

$\frac{1}{x(1-x)}1_{{
0 \leq y < x }} 1_{{0 \leq z < 1-x }}$

[digital valve]

Rion

[pale temple]

Nothing says that y has to be less than 1-z here

So in fact if y > 1-z the pdf is just 0

Well except for that part the remaining computations look sound to me

[spring kernel]

@austere canopy has given 1 rep to @pale temple

[pale temple]

Yeah

[spring kernel]

@austere canopy has given 1 rep to @pale temple

[pale temple]

I don't see anything inherently wrong for that one

[pale temple]

That's the rough sketch yes

Basically you computed the random variable $P(X) = E\left[ 1_{{ Y + Z \leq 1/3}} \vert X \right]$

[digital valve]

Rion

[pale temple]

So now you're taking the expectation of that and use the tower property

$E\left[ E\left[ 1_{{ Y + Z \leq 1/3}} \vert X \right] \right] = E\left[ 1_{{ Y + Z \leq 1/3}} \right]$

[digital valve]

Rion

[pale temple]

To give you an idea of what tower property is:

In other words, note that this kind of integration is only possible due to the fact that X is uniformly distributed in [0, 1]

[pale temple]

It's possible that this integrand might be incorrect btw

You can see that if x is close to 1, z will be almost surely 0 so you shouldn't have to integrate z up to 1/3 - y

So in fact you need to integrate z from 0 to min(1/3 - y, 1 - x)

[pale temple]

Sorry for not seeing that before

As a token of my apology, I'll finish the computations so that you can move on to that part

For $\frac{2}{3} \leq x \leq 1$, with symmetry, you should find that:
\begin{align*}
P(Y + Z \leq 1/3 \vert X = x) &= \int_{0}^{1-x} \left( \int_{0}^{1/3 - z} \frac{1}{x(1-x)}\right)dz \\
&= \frac{1}{x(1-x)} \int_{0}^{1-x} (1/3 - z) dz \\
&= \frac{1}{x(1-x)} \left(\frac{1}{3}(1-x) - \frac{1}{2}(1-x)^2 \right) \\
&= \frac{1}{3x} - \frac{1-x}{2x}
\end{align*}

[digital valve]

Rion

[pale temple]

For $\frac{1}{3} < x < \frac{2}{3}$, you should find that:
\begin{align*}
P(Y + Z \leq 1/3 \vert X = x) &= \int_{0}^{1/3} \left(\int_{0}^{1/3 - y} \frac{1}{x(1-x)} dz \right) dy \\
& = \frac{1}{18x(1-x)}
\end{align*}

[digital valve]

Rion

[spring kernel]

@austere canopy has given 1 rep to @pale temple

[pale temple]

0 to 1/3, 1/3 to 2/3, 2/3 to 1

And then your integration problems should be resolved

[pale temple]

The reasoning is correct at least

I am not sure about the computations of the integrals because I have no pen and paper to verify

So i can only trust you did it correctly

[pale temple]

E[X | Y] is a random variable

Beware of that

E[YZ] = E[E[YZ|X]
= E[ E[Y|X] E[Z|X]]
= E[(X/2)((1-X)/2)]
= (1/4)E[X- X²]
= (1/4)(1/2 - 1/12 - (1/2)² )
= (1/4)(1/6)
= 1/24

Indeed

[spring kernel]

@austere canopy has given 1 rep to @pale temple

[pale temple]

I looked through and I don't see anything blatantly alarming