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Math Help - Some Questions in probability and statistics

  1. #1
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    Some Questions in probability and statistics

    1. Let X~Pois(1) what is the variance of X given x is bigger than 0?
    a. (e-2)/(e-1)
    b. e/(e-1)
    c. 1/(e-2)
    d. e(e-2)/(e-1)^2

    I've tried to use the conditional variance formula but I had problems with calculating the
    probability of x=k and x>0.


    2. A worker calls his wife once a day. the length of each talk is independent of the others, its a random variable which distributes exponentially with a mean of 5.
    One day the boss tells the worker that it is unacceptable and he has 3 more chances, if he calls his wife and talks for more than 10 minutes 3 times he will be fired.
    What is the expected number of days the worker has before he gets fired?
    a. 2e^3
    b. 3e^3
    c. 3e^(-2)
    d. e^2 + e^3

    The parameter of the exponential distribution is 1/5... this is easy.
    but I have no idea what should I do next.
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  2. #2
    Super Member Random Variable's Avatar
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     \displaystyle V(X|X>0) = E(X^2|X>0) - [E(X|X>0)]^2


     \displaystyle  E(X|X>0) =  \frac{\sum_{x=1}^{\infty} xf(x)}{ P(X>0)}

     \displaystyle  E(X^{2}|X>0) =  \frac{\sum_{x=1}^{\infty} x^{2}f(x)}{ P(X>0)}


     \displaystyle f(x) = \frac{1}{x!} \frac{1}{e}

     \displaystyle  P(X>0) = 1 - P(X=0) = 1  - \frac{1}{e} = \frac{e-1}{e}

     \displaystyle \sum_{x=1}^{\infty} xf(x) =\frac{1}{e} \sum^{\infty}_{x=1} \frac{x}{x!} = \frac{1}{e} \ e = 1

     \displaystyle \sum_{x=1}^{\infty} x^{2}f(x) =\frac{1}{e} \sum^{\infty}_{x=1} \frac{x^{2}}{x!} = \frac{1}{e} \ 2e = 2

    so  \displaystyle  E(X|X>0) =  \frac{e}{e-1}

    and  \displaystyle  E(X^{2}|X>0) =  \frac{2e}{e-1}

    then finally  \displaystyle V(X|X>0) = \frac{2e}{e-1} - \frac{e^{2}}{(e-1)^{2}} = \frac{e(e-2)}{(e-1)^{2}}
    Last edited by Random Variable; January 16th 2011 at 11:59 AM. Reason: fixed some errors
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  3. #3
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    Thanks! you've really helped me
    any idea about the second question?
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  4. #4
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    some more questions I had trubles with

    1. Let X1,X2 be independent random variables with equal distribution and CDF of:
    F_X (x) = { x \over x+1}, x \ge 0
    Let Y=X1 \cdot X2 the probability that Y is smaller or equal to 1 is:
    a. 1/4
    b. 1/2
    c. 3/8
    d. none of the above

    I don't even know how to start solving this question....
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  5. #5
    MHF Contributor matheagle's Avatar
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    You start by drawing.
    you want to be "inside" the hyperbola xy<1, where x is x1 and y is x2.
    By differentiation we have the marginals ...

    f(x)=(x+1)^{-2} where x>0

    so the joint density is f(x,y)= (x+1)^{-2} (y+1)^{-2} in the first quadrant

    so you need to integrate \int_0^{\infty}\int_0^{1/x}{dydx\over (x+1)^{2} (y+1)^{2}}
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  6. #6
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    another question i ran into

    Thanks.

    1. The probability that a flip of a coin will get head is a continuous random variable P, with this PDF:
    f_P (p) = \left\{ \begin{array}{1 1} 2 & \quad {1 \over 4} \le p \le {3 \over 4} \\ 0 & \quad \mbox{otherwise} \\ \end{array} \right.}
    Let A be an event in which we got an head in the flip, what is the PDF of P given A?


    another question which I solved but I'm not sure about the answer is:

    2. What is the maximum estimator of
    P_X (x_i \theta) = c(\theta) e^{-\theta x} \quad ,x=0,1,... \quad ,\theta>0
    while c(\theta) is a constant who dose not depend on x:
    a. \log \left( 1 + { n \over \sum_{i=1}^n x_i \right) }
    b. \log \left( 1 + { \sum_{i=1}^n x_i \right) }
    c. \log \left( n + { 1 \over \sum_{i=1}^n x_i \right) }
    d. none of the above.

    I got d...
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  7. #7
    MHF Contributor matheagle's Avatar
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    you left out something
    I assume you want the MLE of theta, you estimate parameters.
    It looks like you must solve for c, I get

    c(\theta)= 1-e^{-\theta} using the geometric series

    so the likelihood function is

    L=  \bigl(1-e^{-\theta}\bigr)^ne^{-\theta\sum x_i}

    Take the log, differentiate, then set equal to zero, flip it over and....THE answer is (a)
    Last edited by matheagle; January 17th 2011 at 04:30 PM.
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  8. #8
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    yea.. i meant the MLE

    Here is another one i solved:

    Let X~U(0,\theta) continuous. A confidence interval to \theta is suggested based on a single observation:
    the confidence level of the confidence interval is:
    a. 1-\alpha
    b. 1-2\alpha
    c. 1-{\alpha \over 2}
    d. you cannot calculate the confidence level from the given data

    I've got b
    Last edited by prosper; January 17th 2011 at 04:35 PM.
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  9. #9
    MHF Contributor matheagle's Avatar
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    I know, but MLE of which paramater, in this case it's obvious that it's theta
    BUT you just don't ask for MLE's
    It's the MLE of .....(an unknown parameter)
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  10. #10
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    I don't quite understand how you got (a) in the MLE question, especially how you found C could you please elaborate?

    I've got another one. and I've added another one to the previous post i made.
    I solved this one too the answer that i got is (a).

    Let W~N(0,0.5). A sensor got signal W when hasnt penetration (H0) and signal W+Theta whan has penetration (alternative H1). Theta>0 constant.

    Got record signal 0.96. Should we reject H0 with Statistical significance of 5% ?
    a. Yes
    b. No
    c. The answer independ with theta
    d. Threre is no way to answer about one record.
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  11. #11
    MHF Contributor matheagle's Avatar
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    A few comments
    (1) You need to create a new thread when asking a different question
    (2) You do need to show some work. Just saying you think the answer is (d), gets old.
    (3) You should hit the thank you link once in awhile.
    Last edited by matheagle; January 18th 2011 at 10:19 PM.
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