Random variables: Total probability, Transformations & CDFs

[danielakkerma]

Hello All!
A recent problem has stuck with me, and I was hoping you could help me resolve it.
Consider the following premise: Let us assume that X \sim \mathcal{U}(-3,3)
(U is the continuous, uniform distribution).
And let the transformation Y be applied thus:
<br /> Y = \left\{<br /> \begin{align*}<br /> X+1, &amp; &amp; -1 \leq X \leq 0 \\<br /> 1-X, &amp; &amp; 0\leq X \leq 1 \\<br /> 0~~~, &amp; &amp; \rm{otherwise}<br /> \end{align*}<br /> \right.<br />
Then one desires to evaluate F_Y(t), Where F(t) is the cumulative dist. func. for Y.
Obviously, the simplest approach would be to find the expression using elementary means -- for example, by plotting the new domain of Y as a function of X.
However, I attempted to obtain the same result by considering the problem from more general principles, particularly, the law of total probability.
I considered the following statement:
<br /> F_Y(t) = \mathbf{P}(Y \leq t)<br />
By LTP:
<br /> \mathbf{P}(Y \leq t) = \sum_{i} P(Y \leq t ~ \mathbf{|} X \in A_i) \cdot P(X \in A_i)<br />
Where {A} is the set of all the regions on which Y is defined, as a function of X. For example, A_1 = [-1,0],A_2 =[0,1], and so forth.
Thus, I would get:
<br /> \mathbf{P}(Y \leq t) = P(Y \leq t ~ \mathbf{|} -1 \leq X \leq 0) \cdot P(-1 \leq X \leq 0) + P(Y \leq t ~ \mathbf{|} 0 \leq X \leq 1) \cdot P(0 \leq X \leq 1) + \\ + P(Y \leq t ~ \mathbf{|} 1 \leq X \leq 3 \cup -3 \leq X \leq -1) \cdot P(1 \leq X \leq 3 \cup -3 \leq X \leq -1)<br />
Then I observe that: P(Y \leq t ~ \mathbf{|} -1 \leq X \leq 0) is merely P(Y \leq t \cap Y = X+1) = P(X+1 \leq t) = F_X(t-1).
This I then apply to all the conditional probabilities(i.e., separating the values of Y according to the constituent Xs(as shown)) and using the CDF for X.
However, I obtain a completely different(and erroneous!) result here, compared with the direct approach(i.e., graphic, and others).
What went wrong?
Is my approach at all correct(or possible/permissible)?
Thank you very much for your attention,
Daniel

 

[Stephen Tashi]

Then I observe that: P(Y \leq t ~ \mathbf{|} -1 \leq X \leq 0) is merely P(Y \leq t \cap Y = X+1) = P(X+1 \leq t) = F_X(t-1).

Does this amount to claiming that P(A|B) = P(A and B) ?

 

[danielakkerma]

Not exactly...

Thanks for your reply.
I meant to say that that event corresponded to:
<br /> P(Y \leq t ~ | ~ -1 \leq X \leq 0) = P(X+1 \leq t)<br />
Since logically, the two are -- or at least should be -- equivalent(one is only possible with the other in tandem).
Is this reasoning invalid?
What is, then, the probability of that conditional statement?
Thanks again,
Daniel

 

[Stephen Tashi]

Suppose t = -1/2.

P(X+1 \le -1/2) = P(X \le -3/2) = \frac{ (-3/2) -(-3)}{6} = 3/12 = 1/4

P(Y \le -1/2 | -1 \le X \le 0) = 0

 

[danielakkerma]

You are, of course, correct!

You're obviously right. I can't believe I didn't detect such a boneheaded mistake, sooner; thank you!
I see I should have written that equality, using the LTP, in this manner:
<br /> P(Y \leq t) = \sum_i P(Y \leq t \cap A_i)
Where again A_i form the domains of Y.
I can therefore get, for one of the subtended regions:
<br /> P(Y \leq t \cap -1\leq X\leq 0) = P(X \leq t-1 \cap -1 \leq X \leq 0)=<br /> \left\{<br /> \begin{align*}<br /> F_X(t-1)-F_X(-1), &amp; &amp; 0 \leq t \leq 1 \\<br /> 1~~~~, &amp;&amp; t&gt;1 \\<br /> 0, &amp;&amp; else<br /> \end{align*}<br /> \right.<br />
But, for the other term:
<br /> P(Y \leq t \cap 0\leq X\leq 1) = P(X \geq 1-t \cap 0 \leq X \leq 1)=<br /> \left\{<br /> \begin{align*}<br /> F_X(1)-F_X(1-t), &amp; &amp; 0 \leq t \leq 1 \\<br /> 1~~~~, &amp;&amp; t&gt;1 \\<br /> 0, &amp;&amp; else<br /> \end{align*}<br /> \right.<br />
Here, it is already evident that when summing these two results(as per the LTP), one would obtain that \lim_{t \to \infty} F_Y(t) = 2 and not 1, which is a fundamental property of the CDF, lost here.
How do I correct this discrepancy?
Thanks,
Daniel

 

[Stephen Tashi]

<br /> P(Y \leq t \cap -1\leq X\leq 0) = P(X \leq t-1 \cap -1 \leq X \leq 0)=<br />

But Y \leq t is not the same event as X \leq t -1. So you can't equate the events Y \leq t \ \cap -1 \leq X \leq 0 and X \leq t-1 \ \cap -1 \leq X \leq 0.

Let Y be a function of the random variable X. Let the sets A_i partition the domain of X.

Then P(Y \leq t) = \sum_{i=1}^n P(Y \le t \ \cap X \in A_i)
= \sum_{i=1}^n P(Y \leq t| X \in A_i) P(X \in A_i).

To compute P(Y \leq t | X \in A_i) you can express the statement that defines Y \leq t as an equivalent statement about X. Then compute P(Y \leq t) as P(Y \leq t \ \cap X \in A_i)/ P(X \in A_i).

For example, in your problem:

\{Y \leq 1/2\} = S = \{-3 \leq X \leq -1\} \cup \{ -1 \leq X \leq -1/2 \} \cup \{ 1/2 \leq X \leq 3 \}

\{Y \leq 1/2\} \cap \{-1 \leq X \leq 0 \} = S \cap \{-1 \leq X \leq 0\} = \{-1 \leq X \leq -1/2\}


The way such examples are usually solved is to express the event Y \le t as an equivalent statement about X being in one of a union of mutually exclusive sets B_i. The sets B_i depend on t. Then the law of total probability is used to compute P(X \in (B_1 \cup B_2\cup...\cup B_n)).

Thinking of Yas a function that maps a set s in the domain of X to a set Y(s) in the domain of Y the way to find P(Y \leq t) is to find the probability of the set Y^{-1}( \{Y \leq t\}) using the distribution of X

 

[danielakkerma]

Now, it's finally clear!

Stephen,
Thanks again for your patient and diligent aid here! it's finally dawned on me(and I'm sorry it has taken so long).
I now see that I should have accounted for the various values Y≤t could take, irrespective of X; and obviously, as you point out, the intersection between Ys and Xs would not -- necessarily -- result in limiting Y itself to any particular domain(as a function of X).

Thinking of Y as a function that maps a set s in the domain of X to a set Y(s) in the domain of Y the way to find P(Y≤t) is to find the probability of the set Y^-1({Y≤t}) using the distribution of X

This is what I have been doing hitherto, but I was hoping I could find a more analytical method to compute these sets, especially, when the transformations are not quite so trivial(and may involve multi-valued inverse solutions).
Still, I quite see now where I was mistaken.
Thanks again for all your help!
Daniel

 

[Stephen Tashi]

but I was hoping I could find a more analytical method to compute these sets, especially, when the transformations are not quite so trivial(and may involve multi-valued inverse solutions).

Don't completely give up on that goal. I don't know what progress can be made, but if you find something, it would be a great service to mathematical humanity. Maybe the cure for the problems of transformations is yet more transformations.