1st Yr Probability: question about marginal and joint pdfs for $3$ uniform continuous independent random variables

Question

Background

I'm trying to improve my understanding of the relationship between marginal and joint pdfs for calculating specific probabilities.

The Problem

$X$, $Y$, $Z$ are independent and uniformly distributed $(0,1)$.

What is $P(X>YZ)$?

My question

The book solution is below, but I'm wondering if I can solve this with the marginal distribution of $X$ alone.

The marginal pdf of $X$ is $f_X(x) = 1$

In theory with $f_X(x)$ I can calculate any probability for $X$. I believe that is the whole point of having a pdf for a random variable.

Therefore:$$P(X>YZ) = \int_{YZ}^{1}dx$$

$$=1-YZ$$

But this definitely isn't the right answer (which as you see below is $3/4$).

The book solution makes complete sense to me.

My question is why can't we get the answer from the marginal pdf of $X$? Shouldn't a marginal pdf for a RV answer all probability statements for that RV?

Thanks for your help and patience!

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Book Solution

Answer 1

To understand why you also need to integrate with respect to $Y$ and $Z$, then think about this sentence:

Evaluating a probability is very similar to evaluating the area/volume of a set.

To be more precise, suppose that you want to find the probability that $$(X,Y,Z) \in V,$$

where $V$ is a particular set in $\mathbb{R}^3$. In you case, $$V = \{(X,Y,Z) : X > YZ, X \in [0,1], Y \in [0,1], Z \in [0,1]\}.$$ Then, the probability associated is:

$$P((X,Y,Z) \in V) = \iiint_V f_{X,Y,Z}\,dx\,dy\,dz,$$

where $f_{X,Y,Z}$ is the joint distribution of $X$, $Y$ and $Z$.

In your specific case, we know that $f_{X,Y,Z} = 1$ for all $(X,Y,Z) \in Q$, where $Q = [0, 1] \times [0, 1] \times [0,1].$

Then, the probability is $$P((X,Y,Z) \in V) = \iiint_V \,dx\,dy\,dz.$$

Notice that, $\iiint_V \,dx\,dy\,dz$ corresponds to the volume of $V$. We can say that the volume of $V$ corresponds to the number of favorable cases. As a final (and obvious) remark, notice that: $$P((X,Y,Z) \in Q) = \iiint_Q \,dx\,dy\,dz$$ is exactly the volume of the cube $Q$. In particular, the probability that $(X,Y,Z) \in Q$, is $1$, and corresponds to the number of possible cases.

Concluding, you need to integrate with respect to all the variables, since calculating a probability is like evaluating the volume of a set.