- #1

danielakkerma

- 231

- 0

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

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

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

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