Results 1 to 13 of 13

Math Help - Desperate! Help! 3 Probability Questions

  1. #1
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56

    Desperate! Help! 3 Probability Questions

    1. Let Y ~ Binomial(n,p) with pmf pY(y).Obtain the value of y for which pY(y) is maximum.

    2. Let Y ~ geometric(p) with pmf pY(y).
    (a) Show that the value p = 1/y maximizes the pmf.
    (b) Obtain E(1/Y ).

    3. Let X be a random variable, and Y = aX + b for non-random values a and b. Using the fact that MY(t) = (e^bt)MX(at) show that V ar(Y ) = (a^2)V ar(X).

    Man...I'm so stressed. The textbook is useless and the prof won't answer questions till all 10 questions are handed in tomorrow. These are the last 3 questions I have.
    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Ruichan View Post
    1. Let Y ~ Binomial(n,p) with pmf pY(y).Obtain the value of y for which pY(y) is maximum.
    This is the mode of the binomial which is:

    \begin{array}{cc}\lfloor (n+1)p \rfloor,&\ \ \ (n+1)p \mbox{ not an integer}\\np \land (n+1)p,&\ \ \ (n+1)p \mbox{ an integer} \end{array}


    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56
    Thanks Captain but I don't understand?
    I've looked through my old textbooks and the current textbook. There's no examples of maximization at all. My prof didn't even talk about it last semester nor this semester (began a week ago).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Ruichan View Post
    Thanks Captain but I don't understand?
    I've looked through my old textbooks and the current textbook. There's no examples of maximization at all. My prof didn't even talk about it last semester nor this semester (began a week ago).
    I'm not surprised it appears to be maximisation with respect to a discrete
    variable which is not something commonly covered in courses.

    In this case we know that the pmf is approximatly symmetric and bell shaped
    so we would expect its maximum to be near the mean, and with a bit of handwaving
    that it is what I gave can be made plausible.

    It makes me think that there may be a mistake in the way the problem has
    been set.

    (the next one is even worse, I can't quite figure out what it wants
    maximised, it can't be to find the k that maximises the pmf for a fixed p
    it must be to find the p that maximises the pmf for fixed k, which at least
    is a problem that is susceptable to the use of calculus)

    RonL
    Last edited by CaptainBlack; January 31st 2007 at 02:28 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Ruichan View Post
    2. Let Y ~ geometric(p) with pmf pY(y).
    (a) Show that the value p = 1/y maximizes the pmf.
    (b) Obtain E(1/Y ).
    a) the pmf for the geometric distribution is:

    f(k,p)= \left\{\begin{array}{cc}(1-p)^{k-1}p,&\ \ \ k \in \{1,2,...\}\\0,&\ \ \ k \notin \{1,2,..\} \end{array}\right.

    so for any k \in \{1,2,...\}, we have:

    \frac{\partial f(k,p)}{\partial p}=(k-1)(1-p)^{k-2}(-1)p+(1-p)^{k-1}

    The maximum of f(k,p) for fixed k occurs when this is equal to zero, or with some rearrangement and cancelling factors of (1-p)^{k-2} is equivalent to:

    -(k-1)p+(1-p)=0,

    which has solution p=1/k.

    (There is a wrinkle that we should worry about briefly, and that is that the maximum could in principle occur when p=1 \mbox{ or } 0, but in this case these are minima for all k>1, so we need not worry about such posibbilities)

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Ruichan View Post
    .

    2. Let Y ~ geometric(p) with pmf pY(y).
    (a) Show that the value p = 1/y maximizes the pmf.
    (b) Obtain E(1/Y ).
    b)

    <br />
E(1/Y)=\sum_{k=1}^{\infty}\frac{(1-p)^{k-1}p}{k}=\frac{p}{1-p}\sum_{k=1}^{\infty}\frac{(1-p)^{k}}{k}<br />

    The last is a standard series and sums to -\ln(p) so:

    <br />
E(1/Y)=\frac{p}{p-1}\ln(p)<br />

    RonL

    (who would not be supprised to find an error in the above)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56
    Quote Originally Posted by CaptainBlack View Post
    b)

    <br />
E(1/Y)=\sum_{k=1}^{\infty}\frac{(1-p)^{k-1}p}{k}=\frac{p}{1-p}\sum_{k=1}^{\infty}\frac{(1-p)^{k}}{k}<br />

    The last is a standard series and sums to -\ln(p) so:

    <br />
E(1/Y)=\frac{p}{p-1}\ln(p)<br />

    RonL

    (who would not be supprised to find an error in the above)
    I attempted question 2. I could get part a) but I didn't know how to get part b). Thank you very much!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2007
    Posts
    42
    Quote Originally Posted by Ruichan View Post
    1. Let Y ~ Binomial(n,p) with pmf pY(y).Obtain the value of y for which pY(y) is maximum.

    2. Let Y ~ geometric(p) with pmf pY(y).
    (a) Show that the value p = 1/y maximizes the pmf.
    (b) Obtain E(1/Y ).

    3. Let X be a random variable, and Y = aX + b for non-random values a and b. Using the fact that MY(t) = (e^bt)MX(at) show that V ar(Y ) = (a^2)V ar(X).

    Man...I'm so stressed. The textbook is useless and the prof won't answer questions till all 10 questions are handed in tomorrow. These are the last 3 questions I have.
    Thanks in advance!

    for discrete case

    if k is the mode

    use
    P(Y=k) >= P(Y= k+1)

    and P(Y=k) >= P(Y=k-1)

    substitute into the pmf and solve the inequalites for k

    some steps are available at mode of the binomial distribution

    some information on derivation of mean, variance ,mode , mgf of binomial distribution is available at derivation of mean, variance ,mode , mgf of binomial distribution
    Last edited by qpmathelp; October 8th 2009 at 07:58 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56
    The only part i could do is part 2a).
    1 and 3, I don't even know how to start.

    Question 3: I know i have to get E(Y) and E(Y^2).
    I just don't know what to do with MY(t) = (e^bt)MX(at).
    How do I differentiate MY(t) = (e^bt)MX(at)?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jan 2007
    Posts
    42
    Quote Originally Posted by Ruichan View Post
    The only part i could do is part 2a).
    1 and 3, I don't even know how to start.

    Question 3: I know i have to get E(Y) and E(Y^2).
    I just don't know what to do with MY(t) = (e^bt)MX(at).
    How do I differentiate MY(t) = (e^bt)MX(at)?
    did you go through the link on my previous post

    a suggestion for question(3) is to expand

    MX(t) = 1 +E(X) t/1! +E(X^2) (t^2) / 2! + ....


    replace t with (at) to get MX(at)

    then use product rule to differentiate MY(t) w.r.t. 't' and put t =0

    hopefully you should get
    E(Y) = a E(X) + b

    diff. MY(t) w.r.t. 't' once more and put t =0 to get E(Y^2)

    and use  Var(Y) = E(Y^2) - {E(Y) }^2
    Last edited by qpmathelp; January 31st 2007 at 09:40 AM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack View Post
    b)

    <br />
E(1/Y)=\sum_{k=1}^{\infty}\frac{(1-p)^{k-1}p}{k}=\frac{p}{1-p}\sum_{k=1}^{\infty}\frac{(1-p)^{k}}{k}<br />

    The last is a standard series and sums to -\ln(p) so:

    <br />
E(1/Y)=\frac{p}{p-1}\ln(p)<br />

    RonL

    (who would not be supprised to find an error in the above)
    For the last part of this we need to obtain the following sum:

    <br />
S(p)=\sum_{k=1}^{\infty}\frac{(1-p)^{k}}{k}<br />

    Consider:

    <br />
f(q)=\sum_{k=0}^{\infty}q^k=\frac{1}{1-q}<br />

    Then consider the following integral:

    <br />
\int_0^q f(x) dx=\int_0^q \left[ \sum_{k=0}^{\infty} x^k \right] dx<br />

    Integrating term by term on the left:

    <br />
\int_0^q f(x) dx= \sum_{k=0}^{\infty}\left[ \int_0^q x^k dx\right] <br />
<br />
= \sum_{k=0}^{\infty} \frac{q^{k+1}}{k+1}= \sum_{k=1}^{\infty} \frac{q^{k}}{k}<br />

    But:

    <br />
\int_0^q f(x) dx=-\ln(1-q)<br />

    So put q=1-p and we have:

    <br />
S(p)=\sum_{k=1}^{\infty}\frac{(1-p)^{k}}{k}=-\ln(p)<br />

    which is what we needed

    RonL
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member Ruichan's Avatar
    Joined
    Oct 2006
    Posts
    56
    Thank you very much Captain.

    I did question 1 this way, not sure if it's correct or not:
    P(Y=y) = pmf of binomial distribution (sorry dunno how to use the math function)
    then i take log P(Y=y)
    Next I differentiate log P(Y=y), then set it to 0.
    Final answer, P=y/n.

    Q3: I got confused at how should I differentiate (e^bt)Mx(at), the Mx(at) part.
    I've figured out already and was able to prove that Var(Y)=a^2Var(X).

    Thank you so much for your help again!
    Till next week....the horror starts again!
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Ruichan View Post
    Thank you very much Captain.

    I did question 1 this way, not sure if it's correct or not:
    P(Y=y) = pmf of binomial distribution (sorry dunno how to use the math function)
    then i take log P(Y=y)
    Next I differentiate log P(Y=y), then set it to 0.
    Final answer, P=y/n.
    You cant differentiate P(y) as y is a discrete variable (only takes integer
    values) which is why the method proposed by qpmathelp while cumbersome
    is the right approach to this.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help please I'm desperate
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: September 11th 2009, 04:47 AM
  2. please help, desperate!
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 14th 2009, 07:06 AM
  3. **urgent: Desperate Probability Question**
    Posted in the Statistics Forum
    Replies: 2
    Last Post: June 3rd 2008, 07:31 PM
  4. In desperate need of some help!!
    Posted in the Algebra Forum
    Replies: 6
    Last Post: September 28th 2007, 05:42 AM
  5. [SOLVED] In need of DESPERATE help
    Posted in the Algebra Forum
    Replies: 9
    Last Post: April 13th 2006, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum