Probability change of random variables question

[tomelwood]

Homework Statement

I am trying to work out how to find the distribution function F$$_{Y}$$ of Y, a random variable given the distribution function F$$_{X}$$ of X and the way that Y is defined given X (see below).

Any pointers to get me started would be brilliant. I have done a similar question to this before, but can't see how to apply what I used there to this one, as previously I didn't start with F$$_{X}$$ , rather with P$$_{X}$$.

Homework Equations

F$$_{X}$$(x) = {0 if x<1
{ $$\frac{x+1}{10}$$ if 1$$\leq$$x<$$\frac{3}{2}$$
{$$\frac{1}{3}$$(x-$$\frac{1}{2}$$ if $$\frac{3}{2}$$$$\leq$$x<$$\frac{5}{2}$$
{1 if x$$\geq$$$$\frac{5}{2}$$


and

Y = {X$$^{2}$$ if X<2
{4 if 2$$\leq$$X<3
{4(4-X) if 3$$\leq$$X<4
{0 if X$$\geq$$4

The Attempt at a Solution

As I said, I'm not entirely sure.
Something has to be done to change it into P$$_{X}$$ I think. But I don't know how to do this.
Or should I differentiate F to get f, the density function?
But then what do I do?

Any help would be greatly appreciated.
Many thanks in advance.
[

 

[mighty2000]

Your Name]



To find the distribution function F_{Y} of Y, we can use the definition of the distribution function as the probability that the random variable Y takes on a value less than or equal to a given number y. In this case, we can break down the definition of Y into four different cases, depending on the value of X.

Case 1: X < 1
In this case, Y = X^2. Therefore, for any y < 0, F_{Y}(y) = P(Y \leq y) = P(X^2 \leq y) = 0, since X^2 is always positive.

Case 2: 1 \leq X < \frac{3}{2}
In this case, Y = X^2. Therefore, for any y < 1, F_{Y}(y) = P(Y \leq y) = P(X^2 \leq y) = P(X \leq \sqrt{y}) = F_{X}(\sqrt{y}).

Case 3: \frac{3}{2} \leq X < \frac{5}{2}
In this case, Y = 4. Therefore, for any y < 4, F_{Y}(y) = P(Y \leq y) = P(4 \leq y) = 0, since Y is always equal to 4 in this range.

Case 4: X \geq \frac{5}{2}
In this case, Y = 4(4-X). Therefore, for any y < 4, F_{Y}(y) = P(Y \leq y) = P(4(4-X) \leq y) = P(X \geq 4-\frac{y}{4}) = 1-F_{X}(4-\frac{y}{4}).

Combining all four cases, we can write the distribution function F_{Y} as:

F_{Y}(y) = {0 if y < 0
{F_{X}(\sqrt{y}) if 0 \leq y < 1
{0 if 1 \leq y < 4
{1-F_{X}(4-\frac{y}{4}) if y \geq 4

I hope this helps you