Probability Distribution function change of variables.

[tomtom690]

Homework Statement

I have been given the distribution function F_X of the random variable X and I am asked to find the distribution function F_Y of Y, another random variable which is defined from X in the following way.

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

The distribution function of X is the following:

F_X (x)={\stackrel{0 if x<1;}{\frac{x+1}{10} if 1 \leq x < \frac{3}{2};}\stackrel{\frac{1}{3}(x-\frac{1}{2}) if \frac{3}{2} \leq x < \frac{5}{2};}{1 if \frac{5}{2} \leq x}
I really have very little idea of how to start this, so any pointers at all would be brilliant. I have done a similar question to this before, but then did not start with the distribution, but rather the probability mass function. And we had to find something different at the end as well.

Homework Equations

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.


After having previewed the post, it is clear that the stacking brackets thing hasn't worked. Please read it as Y having 4 conditions, and the F_X also having 4 conditions. I hope this is clear. The symbols for the imaginary i and the function f together should in fact simply read "if" ie Y= X^2 if X<2.

 

[vela]

Fixed your LaTeX. You can click on the image to see the code I used.

Homework Statement

I have been given the distribution function F_X of the random variable X and I am asked to find the distribution function F_Y of Y, another random variable which is defined from X in the following way.

Y(x) = \left\{<br /> \begin{array}{ll}<br /> x^2 &amp; x&lt;2 \\<br /> 4 &amp; 2 \leq x &lt; 3 \\<br /> 4(4-x) &amp; 3 \leq x &lt; 4 \\<br /> 0 &amp; x \geq 4<br /> \end{array}<br /> \right.

The distribution function of X is the following:

\renewcommand{\arraystretch}{1.3}<br /> F_X(x) = \left\{<br /> \begin{array}{ll}<br /> 0 &amp; x&lt;1 \\<br /> \frac{x+1}{10} &amp; 1 \leq x &lt; \frac{3}{2} \\<br /> \frac{1}{3}(x-\frac{1}{2}) &amp; \frac{3}{2} \leq x &lt; \frac{5}{2} \\<br /> 1 &amp; \frac{5}{2} \leq x<br /> \end{array}<br /> \right.

I really have very little idea of how to start this, so any pointers at all would be brilliant. I have done a similar question to this before, but then did not start with the distribution, but rather the probability mass function. And we had to find something different at the end as well.

Homework Equations

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.


After having previewed the post, it is clear that the stacking brackets thing hasn't worked. Please read it as Y having 4 conditions, and the F_X also having 4 conditions. I hope this is clear. The symbols for the imaginary i and the function f together should in fact simply read "if" ie Y= X^2 if X<2.

F_X(x) is not continuous. Are you sure it's defined correctly?

 

[tomtom690]

Yes. I think that may be the point of the question. Does it mean I can´t do it then? As the next part of the question asks me to split up the F_Y into the discrete and continuous parts- ie F = \lambdaF_1 +(1-\lambda)F_2
Which means it probably isn't continuous in the first place...?
Thanks.

 

[vela]

I just thought it was kind of unusual. I'd probably start by finding the pdf of X.

 

[tomtom690]

OK I´ve managed to get slightly further on this..I think. I'm afraid you'll have to bear with the lack of Latex, as this computer I am at now doesn´t support it for some reason.
So I found the pdf to be
f_X(x) = 0 if x<1
1/10 if 1<=x<3/2
1/3 if 3/2<=x<5/2
0 if x>=5/2

Next, given that Y=g(X) as defined before, I can say that X=sqrt(Y) , X=0, X=(16-Y)/4, and X=0

To find the distr function F_Y(y), which is P(Y<=y) I can say that working in the interval [0,4) , F_Y(y) = P(0<x<=sqrt(y))+P(x>=(16-y)/4)

Now to do this I know you need to integrate, and then do some adding, because on this interval the pdf of x is both 1/10 and 1/3, but I am unsure of the limits of integration...
I think it should be "integrate '1/10' between 1 and (sqrt(y))" + "integrate '1/10' between (16-y)/4 and 5/2" and then the same for 1/3.

Then F_Y(y) = 0 if y<0
"the answer to the above" if 0<=y<4
1 if y>=4

Is this correct? Or have I made some stupid mistake somewhere...?

However, next, I have to split it into the discrete and continuous dist functions, as I mentioned earlier, however using the dist above, I get P_F({0}) = -13/20 which I'm pretty sure is not allowed...

And then I don't really know where to go from there.

Thanks again, and sorry for such a long waffley reply!

 

[vela]

OK I´ve managed to get slightly further on this..I think. I'm afraid you'll have to bear with the lack of Latex, as this computer I am at now doesn´t support it for some reason.
So I found the pdf to be
f_X(x) = 0 if x<1
1/10 if 1<=x<3/2
1/3 if 3/2<=x<5/2
0 if x>=5/2

That's not complete. If you integrate that function, it doesn't come out to be 1. You need to account for the discontinuities in F_X.

 

[tomtom690]

Sorry I have no idea how to do that! I'm not sure I've ever been taught it. Though obviously it is necessary to do this question! Could you explain how, please, or point me in the direction of somewhere that can? Thanks very much!

 

[vela]

Hint: The discontinuities have to do with the discrete probabilities.