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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!

Book Solution

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HJ_beginner
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  • Whether the event $X>YZ$ occurs or not depends not only on $X$ but also on $Y$ and $Z,$ so you can't get the answer only from the distribution of $X.$ You're looking for a number here, not for a random variable, and $YZ$ is a random variable. $\qquad$ – Michael Hardy Feb 19 '18 at 21:35
  • Thanks for your help and patience Michael. I am a beginner. Is it fair to say that a marginal pdf will answer any probability statement about $X$ except if other RVs are involved? Then can I say that knowing a marginal pdf is not "that powerful" since we often want to know probability calculations for more than one RV at a time for jointly distributed RVs. Also, how did you know that we are looking for a number here? My intuition for $P(X>YZ)$ is actually that the probability should be a function of $y$ and $z$ and not just $3/4$ no matter what $x$, $y$, and $z$ are. – HJ_beginner Feb 19 '18 at 21:45

1 Answers1

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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.

the_candyman
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  • Thanks for answer - what you wrote makes sense. But can you expand more on my specific question about why using the marginal distribution of $X$ alone doesn't work? You did touch on this by talking about volume versus 1 dimensional line. I get it that to do a volume you can't just use one RV. But then what is the point of knowing the marginals then? What application is there of knowing the marginals for calculating probabilities? Why is $f_X(x)$ alone not enough to calculate a probability statement on $X$ besides just saying it must be a volume. Can you be more specific because I am a newb. – HJ_beginner Feb 19 '18 at 23:38
  • @HJ_beginner You are broading your question. Anyway, you need the marginals when, for example, you are concerned to find probabilities that are related only to a part of the given random variables. For example, if $V={(x,y,z) : x > 0.5}$, then you can use just the marginal of $x$. Another important fact about the marginals is that, when you have $N$ independent random variables, then the joint pdf is equal to product of all the marginals. – the_candyman Feb 19 '18 at 23:41
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    So with the marginal pdf of $X$... you can calculate specific probabilities for statements that don't involve any other RVs. For example, the set $V$ you mentioned. However if you have a $V_2 = {(x,y,z) : x > y}$ then you cannot use the marginal alone to get a specific calculation... is this correct? – HJ_beginner Feb 19 '18 at 23:47
  • @HJ_beginner exactly. Or for the case of independent rv, you can use the marginals to find the joint pdf. – the_candyman Feb 19 '18 at 23:49
  • Hmm okay thanks. I do appreciate your help. I will re-read your answer a few times, it takes a few days for things to sink into my mind. – HJ_beginner Feb 19 '18 at 23:50
  • I have downvoted this because you speak of the probability that $(x,y,z)$ is in some set rather than of the probability that $(X,Y,Z)$ is in some set, so you're using lower-case $x,y,z$ as names of random variables, but in the next breath you use those same letters from the variables of integration. There are good reasons why one writes, for example, $\displaystyle \iiint_{\mathbb R^3} f_{X,Y,Z}(x,y,z) , dx,dy,dz,$ with capitals in certain places and lower-case letters in certain places. Notice that $f_X(3)$ and $f_Y(3)$ are two different things, and $\Pr(X\le x)$ is a$,\ldots\qquad$ – Michael Hardy Feb 20 '18 at 03:08
  • $\ldots,$function of $x$ (not of $X$), and the expression $\Pr(X\le x)$ would be incomprehensible if $x$ is the same thing as $X. \qquad$ – Michael Hardy Feb 20 '18 at 03:10
  • @MichaelHardy You are right, thanks for this remark. I fixed it. – the_candyman Feb 20 '18 at 08:02