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Thread: Probability System, discrete dist.

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    Re: Probability System, discrete dist.




    because the distribution is 0 1 3
    so if you make a chart of X vs. Y you have 0 1 3 on both axes and you cannot get five out of those two when you fill out the chart

    \begin{tabular}{|c||c|c|c|} \hline
& 0 & 1 & 3 \\ \hline \hline
0 &  &  &  \\ \hline
1 &  &  &   \\ \hline
3 &  &  & \\ \hline
\end{tabular}

    and the marginal probabilities are 0.1, 0.5, 0.4
    and you can combine different combinations from 0-3 to get your pmf

    make sense?

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    Re: Probability System, discrete dist.

    Yes that make sense to me, if the support is \{0, 1, 3\} instead of \{0, 1, 2\}

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    Re: Probability System, discrete dist.

    so from that i created the PMF, not pdf since this is a discrete distribution

    i just don't know how im suppose to get the cdf from the pmf i had posted... hints?

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    Re: Probability System, discrete dist.

    for the cdf should I draw a graph of the pmf and then integrate somehow?
    its so much easier when its converting a pdf to cdf, im not remembering how to do it for a pmf..
    help.

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    Re: Probability System, discrete dist.

    The cdf still has the same definition. It's just that instead of integrating from negative infinity up to a point 'x' you're summing up to the point 'x'. If you look at it from a measure theory standpoint it's essentially the same thing.

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    Re: Probability System, discrete dist.

    Yes. Just need to be careful about when to jump.

    Once you have your pmf

    \Pr\{X = x\}, x = 0, 1, 2, 3, 4, 6

    Then you can get, for example

    \Pr\{X \leq x\} = 0 ~ \forall x < 0

    \Pr\{X \leq x\} = \Pr\{X = 0\} ~ \forall x \in [0, 1)

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    Re: Probability System, discrete dist.

    so for 0<=x<=1 it would be 0.01
    and for 1<=x<=2 it would be 0.1 or 0.11? i was confused which numbers to add or subtract for the different probabilities in the pmf

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    Re: Probability System, discrete dist.

    Be careful... you are dealing with discrete random variable, so need to pay
    extra attention to the sign \leq here.

    I give you 1 more example:

    \begin{array}{|l|l|} \hline
x & F_X(x) \\ \hline
x < 0 & 0 \\
0 \leq x < 1 & \Pr\{X = 0\} \\
1 \leq x < 2 & \Pr\{X = 0\} + \Pr\{X = 1\} \\
2 \leq x < 3 & \Pr\{X = 0\} + \Pr\{X = 1\} + \Pr\{X = 2\} \\ \hline
\end{array}

    So you can complete the remaining part yourself.

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    Re: Probability System, discrete dist.

    ahh i had a feeling thats how it was suppose to be. because if i graph the pmf and draw a line diagonally where 2<=x<3 it would include the point x=0,1,2 underneath it, since the cdf is the area under the curve of the pmf.

    thanks so much for the clarification

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    Re: Probability System, discrete dist.

    one more question
    if this was instead of a discrete distribution but a uniform distribution on interval [0,2]

    i got the pdf to be {0.25 if 0<=x<=2, 0 if otherwise}

    now when I tried to get the cdf I had trouble with 2<=x<4
    here is what I have so far for the cdf
    {0 if x<=0, (1/8)x^2 if 0<=x<2, 1 if x>=4}
    this is a joint probability so I made a 2x2 square for the pdf and tried to find the area of the triangle for each interval and that was how I got (1/8)x^2, but I am unsure how to get the cdf when 2<=x<4
    any hints?

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    Re: Probability System, discrete dist.

    this is a joint probability so I made a 2x2 square for the pdf and tried to find the area of the triangle for each interval
    You are talking about the joint CDF or just the univariate case?
    any thing I missed?

    If you are talking about the univariate case, then

    if this was instead of a discrete distribution but a uniform distribution on interval [0,2]

    i got the pdf to be {0.25 if 0<=x<=2, 0 if otherwise}
    now when I tried to get the cdf I had trouble with 2<=x<4
    here is what I have so far for the cdf
    {0 if x<=0, (1/8)x^2 if 0<=x<2, 1 if x>=4}
    are incorrect.

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    Re: Probability System, discrete dist.

    im almost 100% sure i have the pdf right, FOR UNIFORM distribution on [0,2]

    but any hints on the cdf for that? ive been trying to solve it visually with a graph of the interval which is a 2x2 square.

    and i thought i was on a roll

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    Re: Probability System, discrete dist.

    Quote Originally Posted by gator78 View Post
    im almost 100% sure i have the pdf right, FOR UNIFORM distribution on [0,2]
    Shouldn't it be 0.5 instead?

    Or you are talking about two independent uniform random variables
    have a joint density on a 2 by 2 square? So in that case will be 0.25?

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    Re: Probability System, discrete dist.


    sorry i was talking about joint independent variables since the system X=X1+X2

    Im understanding this a lot better now. really appreciate the help

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