Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Random variables: Total probability, Transformations & CDFs

  1. Mar 1, 2014 #1
    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 [tex] X \sim \mathcal{U}(-3,3) [/tex]
    (U is the continuous, uniform distribution).
    And let the transformation Y be applied thus:
    [tex]
    Y = \left\{
    \begin{align*}
    X+1, & & -1 \leq X \leq 0 \\
    1-X, & & 0\leq X \leq 1 \\
    0~~~, & & \rm{otherwise}
    \end{align*}
    \right.
    [/tex]
    Then one desires to evaluate [itex] F_Y(t) [/itex], 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:
    [tex]
    F_Y(t) = \mathbf{P}(Y \leq t)
    [/tex]
    By LTP:
    [tex]
    \mathbf{P}(Y \leq t) = \sum_{i} P(Y \leq t ~ \mathbf{|} X \in A_i) \cdot P(X \in A_i)
    [/tex]
    Where [itex]{A}[/itex] is the set of all the regions on which Y is defined, as a function of X. For example, [itex]A_1 = [-1,0][/itex],[itex]A_2 =[0,1][/itex], and so forth.
    Thus, I would get:
    [tex]
    \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)
    [/tex]
    Then I observe that: [itex]P(Y \leq t ~ \mathbf{|} -1 \leq X \leq 0)[/itex] is merely [itex]P(Y \leq t \cap Y = X+1) = P(X+1 \leq t) = F_X(t-1)[/itex].
    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
     
    Last edited: Mar 1, 2014
  2. jcsd
  3. Mar 1, 2014 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    Does this amount to claiming that P(A|B) = P(A and B) ?
     
  4. Mar 1, 2014 #3
    Not exactly...

    Thanks for your reply.
    I meant to say that that event corresponded to:
    [tex]
    P(Y \leq t ~ | ~ -1 \leq X \leq 0) = P(X+1 \leq t)
    [/tex]
    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
     
  5. Mar 1, 2014 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    Suppose [itex] t = -1/2 [/itex].

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

    [itex] P(Y \le -1/2 | -1 \le X \le 0) = 0 [/itex]
     
  6. Mar 1, 2014 #5
    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:
    [tex]
    P(Y \leq t) = \sum_i P(Y \leq t \cap A_i) [/tex]
    Where again [itex] A_i [/itex] form the domains of Y.
    I can therefore get, for one of the subtended regions:
    [tex]
    P(Y \leq t \cap -1\leq X\leq 0) = P(X \leq t-1 \cap -1 \leq X \leq 0)=
    \left\{
    \begin{align*}
    F_X(t-1)-F_X(-1), & & 0 \leq t \leq 1 \\
    1~~~~, && t>1 \\
    0, && else
    \end{align*}
    \right.
    [/tex]
    But, for the other term:
    [tex]
    P(Y \leq t \cap 0\leq X\leq 1) = P(X \geq 1-t \cap 0 \leq X \leq 1)=
    \left\{
    \begin{align*}
    F_X(1)-F_X(1-t), & & 0 \leq t \leq 1 \\
    1~~~~, && t>1 \\
    0, && else
    \end{align*}
    \right.
    [/tex]
    Here, it is already evident that when summing these two results(as per the LTP), one would obtain that [tex]\lim_{t \to \infty} F_Y(t) = 2 [/tex] and not 1, which is a fundamental property of the CDF, lost here.
    How do I correct this discrepancy?
    Thanks,
    Daniel
     
  7. Mar 1, 2014 #6

    Stephen Tashi

    User Avatar
    Science Advisor

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

    Let [itex] Y [/itex] be a function of the random variable [itex] X [/itex]. Let the sets [itex] A_i [/itex] partition the domain of [itex] X [/itex].

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

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

    For example, in your problem:

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

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


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

    Thinking of [itex]Y [/itex]as a function that maps a set [itex] s [/itex] in the domain of [itex]X [/itex] to a set [itex] Y(s) [/itex] in the domain of [itex] Y [/itex] the way to find [itex] P(Y \leq t) [/itex] is to find the probability of the set [itex] Y^{-1}( \{Y \leq t\}) [/itex] using the distribution of [itex] X [/itex]
     
  8. Mar 2, 2014 #7
    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).
    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
     
  9. Mar 2, 2014 #8

    Stephen Tashi

    User Avatar
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Random variables: Total probability, Transformations & CDFs
Loading...