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How do you show the number of subsets of a set is 2^n?

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  • #31
    Originally posted by Zeigy View Post
    Help please. I've really gotten sick and tired of working with integrals of the form e^[f(x)]. You always hope they structure it in such a way that you don't have to go reviewing the rules of integration. Thank God I'm on the last chapter.

    I'm working with distributions of functions of two random variables.

    The question states that for the joint distribution function of a random variable 'S' f(s) = 6[e^(-2s) - e^(-3s)]

    S = X + Y

    Find P(X + Y ≤ 1.5). Hint: Find P(X + Y > 1.5) first.

    What I have done so far is to draw the region for X + Y > 1.5.

    I was trying to double integrate f(s) with respect to x and y for y = 1.5 to ∞ and x = 1.5 to (1.5 - y).

    Am I going right so far? Do I have to integrate that complex function after substituting the s for x and y?
    f(s) is not a joint distribution. I would like to know where you got a problem that tells you that it is.

    A distribution of the sum of random variables is not the same thing as a joint distribution.

    You have the distribution of X+Y. You want to find when X+Y is less than or equal to 1.5. Integrate f(s) from 1.5 to infinity and subtract from one.

    I get 0.8729 when i do this

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    • #32
      X has the marginal distribution function f(x) = 2e^(-2x), x ≥ 0

      Y has the marginal distribution function f(y) = 3e^(-3y), y ≥ 0

      They are independent random variables so I used the formula in an earlier problem to determine how the two interacted to produce the joint distribution f(s) for S = X + Y.

      i.e. f(s) = ∫[-∞ to ∞] f(x) · f(s-x) dx

      The answer to that is

      f(s) = 6[e^(-2s) - e^(-3s)]


      The answer to my previous question is 0.95833
      Last edited by Zeigy; July 9 2011, 07:00 PM. Reason: Clarity

      Comment


      • #33
        Originally posted by Zeigy View Post
        X has the marginal distribution function f(x) = 2e^(-2x), x ≥ 0

        Y has the marginal distribution function f(y) = 3e^(-3y), y ≥ 0

        They are independent random variables so I used the formula in an earlier problem to determine how the two interacted to produce the joint distribution f(s) for S = X + Y.

        i.e. f(s) = ∫[-∞ to ∞] f(x) · f(s-x) dx

        The answer to that is

        f(s) = 6[e^(-2s) - e^(-3s)]


        The answer to my previous question is 0.95833
        A tip: When you ask a question, ask the whole question. Don't present the work you've done half way through the problem as given information from the problem.

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        • #34
          I explained to you that S was a joint distribution, Maestro. All I did by posting "the whole question" as you put it was to show you S was a joint distribution since you didn't want to take my word for it.

          Now answer the question because you are pissing me off.

          Comment


          • #35
            Originally posted by Zeigy View Post
            Now answer the question because you are pissing me off.
            Seriously? What are you five?

            1.) Learn to find a joint density function when presented with two independent univariate density functions. (Both of which you did not provide when you first posted the question).

            2.) Go ahead and keep believing that a function of one variable can be a joint density function.

            3.) Don't worry, I won't be around to piss you off anymore by explaining these things to you.

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            • #36
              To all: if you can't communicate nicely, don't communicate at all. Name calling isn't allowed here. Think about what's probably acceptable in the workplace - that's probably a good guide for what's acceptable here.
              "You better get to living, because dying's a pain in the ***." - Frank Sinatra

              http://www.hockeybuzz.com/blogger_ar...blogger_id=174 - where I talk about the Blues and the NHL.

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              • #37
                My apologies.

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                • #38
                  Am I integrating this right?

                  ∫(α to C) (x - α) · (1/C) dx = C/2 - 1 - α²/2C + α/C

                  The book keeps saying the answer is:

                  (C - α)²/2C

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                  • #39
                    Can someone explain to me the minimum and maximum of any two independent random variables? The textbook isn't too clear. You should probably start with what is a minimum function and what is a maximum function.

                    Comment


                    • #40
                      Originally posted by Zeigy View Post
                      Can someone explain to me the minimum and maximum of any two independent random variables? The textbook isn't too clear. You should probably start with what is a minimum function and what is a maximum function.
                      Why don't you start with defining the min and max functions?

                      Comment


                      • #41
                        Originally posted by Zeigy View Post
                        Am I integrating this right?

                        ∫(α to C) (x - α) · (1/C) dx = C/2 - 1 - α²/2C + α/C

                        The book keeps saying the answer is:

                        (C - α)²/2C
                        Show your work.

                        Comment


                        • #42
                          The maximum and minimum of a function occur for critical values on the x-axis. These are the points at which the derivatives of the function when evaluated at the critical value is equal to zero or the graph reaches a turning point or the gradient is flat.

                          Comment


                          • #43
                            Originally posted by Zeigy View Post
                            The maximum and minimum of a function occur for critical values on the x-axis. These are the points at which the derivatives of the function when evaluated at the critical value is equal to zero or the graph reaches a turning point or the gradient is flat.
                            1) What about points of inflection? What about on a closed interval, etc.

                            2) I meant what's the definition of maximum not maximum of a function i.e. define for me max{X,Y}

                            Comment


                            • #44
                              There is a difference? That's where my problem lies then. Okay. Let me do more revision.

                              Comment


                              • #45
                                Originally posted by Zeigy View Post
                                There is a difference? That's where my problem lies then. Okay. Let me do more revision.
                                Well conceptually max{X,Y} takes in 2 variables and spits one out. So you can think of it as f: R^2 -> R or f(a,b) = c or however you want [it can actually take in more than 2 variables but that's the case I presented]. So think about it max{X,Y} = blah when something is true.

                                1) What's blah?

                                2) When what is true?

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