Finding the Probability with Moment Generating Function












1












$begingroup$


This is a question in a book, but I don't know how to solve it.



Problem



if $X$ is a random variable with $M_X(t)=e^{2t+2t^2}$ then find $P{|X -frac{1}{2}|>2}$.

Question didn't say anything about $X$ distribution (discrete or continuous)



My Solution



Find exact value



for solve this I first try this fact that $$ M_X(ln(t))= psi_X(t) $$

and
$$frac{partial^k}{partial t^k}psi_X(t)|_{t=0}=k!f_X(k)$$



if I calculate $psi_X(t)$, it is:
$$psi_X(t)=t^2e^{2(lnt)^2}$$



but because of $t^2$ there is no easy way to calculate $psi_X(t)$ and if I put $k=0$, I couldn't find $f_x(0)$ because of $ln(0)$!!

Besides, I can only find $f(0), f(1), ...$ but I don't think that our distribution is discrete.



Find upper/lower



I want to know, Is there any way to find upper/lower for probability?



We find upper/lower for $P{|X -frac{1}{2}|>2}$ with Chebyshev inequality. We know that $mu=2$ and $sigma=2$ from $M_X(t)$ Derivations, but I couldn't convert given probability to this form:
$$P{|X-mu|<ksigma} ge 1-frac{1}{k^2}$$

Is it possible to show me how to convert it to above form?



Is it possible to show me something for solve this problem? Thanks.






statistics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    MGFs of some standard distributions have a known form. In all those cases, you can identify the distribution just by looking at the MGF. The one in your question is a known MGF, that of a Normal distribution.
    $endgroup$
    – StubbornAtom
    Dec 18 '18 at 19:34










  • $begingroup$
    wow. it is great. I don't know this. however is there any other way to solve it? because this question is related to Chebyshev and Moment Generating Function not especial distributions? (we don't have normal table in question)
    $endgroup$
    – Amin
    Dec 18 '18 at 19:44










  • $begingroup$
    @Amin When you say Chebyshev, did you mean Chebyshev inequality? In that case, are you trying to calculate the probability precisely or upper bound it? Note that you can find the first and second moments of this distribution by expanding out the MGF (hint: write the series expansion of $e^x$) and collecting the terms with $t$ and $t^2$. After that, it will be easy to apply the inequality.
    $endgroup$
    – Aditya Dua
    Dec 18 '18 at 20:10










  • $begingroup$
    Yes. I mean Chebyshev inequality?. I did this in my question, at the bottom of question. however I couln't convert given probability to Chebyshev inequality form. Is it possible to help me?
    $endgroup$
    – Amin
    Dec 18 '18 at 20:13






  • 1




    $begingroup$
    The question seems to be asking for the exact value of $P(|X -frac{1}{2}|>2)$ so is unlikely to need the Chebyshev inequality. But you can find the mean and variance from the Moment Generating function
    $endgroup$
    – Henry
    Dec 18 '18 at 23:42
















1












$begingroup$


This is a question in a book, but I don't know how to solve it.



Problem



if $X$ is a random variable with $M_X(t)=e^{2t+2t^2}$ then find $P{|X -frac{1}{2}|>2}$.

Question didn't say anything about $X$ distribution (discrete or continuous)



My Solution



Find exact value



for solve this I first try this fact that $$ M_X(ln(t))= psi_X(t) $$

and
$$frac{partial^k}{partial t^k}psi_X(t)|_{t=0}=k!f_X(k)$$



if I calculate $psi_X(t)$, it is:
$$psi_X(t)=t^2e^{2(lnt)^2}$$



but because of $t^2$ there is no easy way to calculate $psi_X(t)$ and if I put $k=0$, I couldn't find $f_x(0)$ because of $ln(0)$!!

Besides, I can only find $f(0), f(1), ...$ but I don't think that our distribution is discrete.



Find upper/lower



I want to know, Is there any way to find upper/lower for probability?



We find upper/lower for $P{|X -frac{1}{2}|>2}$ with Chebyshev inequality. We know that $mu=2$ and $sigma=2$ from $M_X(t)$ Derivations, but I couldn't convert given probability to this form:
$$P{|X-mu|<ksigma} ge 1-frac{1}{k^2}$$

Is it possible to show me how to convert it to above form?



Is it possible to show me something for solve this problem? Thanks.






statistics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    MGFs of some standard distributions have a known form. In all those cases, you can identify the distribution just by looking at the MGF. The one in your question is a known MGF, that of a Normal distribution.
    $endgroup$
    – StubbornAtom
    Dec 18 '18 at 19:34










  • $begingroup$
    wow. it is great. I don't know this. however is there any other way to solve it? because this question is related to Chebyshev and Moment Generating Function not especial distributions? (we don't have normal table in question)
    $endgroup$
    – Amin
    Dec 18 '18 at 19:44










  • $begingroup$
    @Amin When you say Chebyshev, did you mean Chebyshev inequality? In that case, are you trying to calculate the probability precisely or upper bound it? Note that you can find the first and second moments of this distribution by expanding out the MGF (hint: write the series expansion of $e^x$) and collecting the terms with $t$ and $t^2$. After that, it will be easy to apply the inequality.
    $endgroup$
    – Aditya Dua
    Dec 18 '18 at 20:10










  • $begingroup$
    Yes. I mean Chebyshev inequality?. I did this in my question, at the bottom of question. however I couln't convert given probability to Chebyshev inequality form. Is it possible to help me?
    $endgroup$
    – Amin
    Dec 18 '18 at 20:13






  • 1




    $begingroup$
    The question seems to be asking for the exact value of $P(|X -frac{1}{2}|>2)$ so is unlikely to need the Chebyshev inequality. But you can find the mean and variance from the Moment Generating function
    $endgroup$
    – Henry
    Dec 18 '18 at 23:42














1












1








1





$begingroup$


This is a question in a book, but I don't know how to solve it.



Problem



if $X$ is a random variable with $M_X(t)=e^{2t+2t^2}$ then find $P{|X -frac{1}{2}|>2}$.

Question didn't say anything about $X$ distribution (discrete or continuous)



My Solution



Find exact value



for solve this I first try this fact that $$ M_X(ln(t))= psi_X(t) $$

and
$$frac{partial^k}{partial t^k}psi_X(t)|_{t=0}=k!f_X(k)$$



if I calculate $psi_X(t)$, it is:
$$psi_X(t)=t^2e^{2(lnt)^2}$$



but because of $t^2$ there is no easy way to calculate $psi_X(t)$ and if I put $k=0$, I couldn't find $f_x(0)$ because of $ln(0)$!!

Besides, I can only find $f(0), f(1), ...$ but I don't think that our distribution is discrete.



Find upper/lower



I want to know, Is there any way to find upper/lower for probability?



We find upper/lower for $P{|X -frac{1}{2}|>2}$ with Chebyshev inequality. We know that $mu=2$ and $sigma=2$ from $M_X(t)$ Derivations, but I couldn't convert given probability to this form:
$$P{|X-mu|<ksigma} ge 1-frac{1}{k^2}$$

Is it possible to show me how to convert it to above form?



Is it possible to show me something for solve this problem? Thanks.






statistics







share|cite|improve this question











$endgroup$




This is a question in a book, but I don't know how to solve it.



Problem



if $X$ is a random variable with $M_X(t)=e^{2t+2t^2}$ then find $P{|X -frac{1}{2}|>2}$.

Question didn't say anything about $X$ distribution (discrete or continuous)



My Solution



Find exact value



for solve this I first try this fact that $$ M_X(ln(t))= psi_X(t) $$

and
$$frac{partial^k}{partial t^k}psi_X(t)|_{t=0}=k!f_X(k)$$



if I calculate $psi_X(t)$, it is:
$$psi_X(t)=t^2e^{2(lnt)^2}$$



but because of $t^2$ there is no easy way to calculate $psi_X(t)$ and if I put $k=0$, I couldn't find $f_x(0)$ because of $ln(0)$!!

Besides, I can only find $f(0), f(1), ...$ but I don't think that our distribution is discrete.



Find upper/lower



I want to know, Is there any way to find upper/lower for probability?



We find upper/lower for $P{|X -frac{1}{2}|>2}$ with Chebyshev inequality. We know that $mu=2$ and $sigma=2$ from $M_X(t)$ Derivations, but I couldn't convert given probability to this form:
$$P{|X-mu|<ksigma} ge 1-frac{1}{k^2}$$

Is it possible to show me how to convert it to above form?



Is it possible to show me something for solve this problem? Thanks.






statistics




statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 2:59







Amin

















asked Dec 18 '18 at 19:17









AminAmin

355110




355110








  • 2




    $begingroup$
    MGFs of some standard distributions have a known form. In all those cases, you can identify the distribution just by looking at the MGF. The one in your question is a known MGF, that of a Normal distribution.
    $endgroup$
    – StubbornAtom
    Dec 18 '18 at 19:34










  • $begingroup$
    wow. it is great. I don't know this. however is there any other way to solve it? because this question is related to Chebyshev and Moment Generating Function not especial distributions? (we don't have normal table in question)
    $endgroup$
    – Amin
    Dec 18 '18 at 19:44










  • $begingroup$
    @Amin When you say Chebyshev, did you mean Chebyshev inequality? In that case, are you trying to calculate the probability precisely or upper bound it? Note that you can find the first and second moments of this distribution by expanding out the MGF (hint: write the series expansion of $e^x$) and collecting the terms with $t$ and $t^2$. After that, it will be easy to apply the inequality.
    $endgroup$
    – Aditya Dua
    Dec 18 '18 at 20:10










  • $begingroup$
    Yes. I mean Chebyshev inequality?. I did this in my question, at the bottom of question. however I couln't convert given probability to Chebyshev inequality form. Is it possible to help me?
    $endgroup$
    – Amin
    Dec 18 '18 at 20:13






  • 1




    $begingroup$
    The question seems to be asking for the exact value of $P(|X -frac{1}{2}|>2)$ so is unlikely to need the Chebyshev inequality. But you can find the mean and variance from the Moment Generating function
    $endgroup$
    – Henry
    Dec 18 '18 at 23:42














  • 2




    $begingroup$
    MGFs of some standard distributions have a known form. In all those cases, you can identify the distribution just by looking at the MGF. The one in your question is a known MGF, that of a Normal distribution.
    $endgroup$
    – StubbornAtom
    Dec 18 '18 at 19:34










  • $begingroup$
    wow. it is great. I don't know this. however is there any other way to solve it? because this question is related to Chebyshev and Moment Generating Function not especial distributions? (we don't have normal table in question)
    $endgroup$
    – Amin
    Dec 18 '18 at 19:44










  • $begingroup$
    @Amin When you say Chebyshev, did you mean Chebyshev inequality? In that case, are you trying to calculate the probability precisely or upper bound it? Note that you can find the first and second moments of this distribution by expanding out the MGF (hint: write the series expansion of $e^x$) and collecting the terms with $t$ and $t^2$. After that, it will be easy to apply the inequality.
    $endgroup$
    – Aditya Dua
    Dec 18 '18 at 20:10










  • $begingroup$
    Yes. I mean Chebyshev inequality?. I did this in my question, at the bottom of question. however I couln't convert given probability to Chebyshev inequality form. Is it possible to help me?
    $endgroup$
    – Amin
    Dec 18 '18 at 20:13






  • 1




    $begingroup$
    The question seems to be asking for the exact value of $P(|X -frac{1}{2}|>2)$ so is unlikely to need the Chebyshev inequality. But you can find the mean and variance from the Moment Generating function
    $endgroup$
    – Henry
    Dec 18 '18 at 23:42








2




2




$begingroup$
MGFs of some standard distributions have a known form. In all those cases, you can identify the distribution just by looking at the MGF. The one in your question is a known MGF, that of a Normal distribution.
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:34




$begingroup$
MGFs of some standard distributions have a known form. In all those cases, you can identify the distribution just by looking at the MGF. The one in your question is a known MGF, that of a Normal distribution.
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:34












$begingroup$
wow. it is great. I don't know this. however is there any other way to solve it? because this question is related to Chebyshev and Moment Generating Function not especial distributions? (we don't have normal table in question)
$endgroup$
– Amin
Dec 18 '18 at 19:44




$begingroup$
wow. it is great. I don't know this. however is there any other way to solve it? because this question is related to Chebyshev and Moment Generating Function not especial distributions? (we don't have normal table in question)
$endgroup$
– Amin
Dec 18 '18 at 19:44












$begingroup$
@Amin When you say Chebyshev, did you mean Chebyshev inequality? In that case, are you trying to calculate the probability precisely or upper bound it? Note that you can find the first and second moments of this distribution by expanding out the MGF (hint: write the series expansion of $e^x$) and collecting the terms with $t$ and $t^2$. After that, it will be easy to apply the inequality.
$endgroup$
– Aditya Dua
Dec 18 '18 at 20:10




$begingroup$
@Amin When you say Chebyshev, did you mean Chebyshev inequality? In that case, are you trying to calculate the probability precisely or upper bound it? Note that you can find the first and second moments of this distribution by expanding out the MGF (hint: write the series expansion of $e^x$) and collecting the terms with $t$ and $t^2$. After that, it will be easy to apply the inequality.
$endgroup$
– Aditya Dua
Dec 18 '18 at 20:10












$begingroup$
Yes. I mean Chebyshev inequality?. I did this in my question, at the bottom of question. however I couln't convert given probability to Chebyshev inequality form. Is it possible to help me?
$endgroup$
– Amin
Dec 18 '18 at 20:13




$begingroup$
Yes. I mean Chebyshev inequality?. I did this in my question, at the bottom of question. however I couln't convert given probability to Chebyshev inequality form. Is it possible to help me?
$endgroup$
– Amin
Dec 18 '18 at 20:13




1




1




$begingroup$
The question seems to be asking for the exact value of $P(|X -frac{1}{2}|>2)$ so is unlikely to need the Chebyshev inequality. But you can find the mean and variance from the Moment Generating function
$endgroup$
– Henry
Dec 18 '18 at 23:42




$begingroup$
The question seems to be asking for the exact value of $P(|X -frac{1}{2}|>2)$ so is unlikely to need the Chebyshev inequality. But you can find the mean and variance from the Moment Generating function
$endgroup$
– Henry
Dec 18 '18 at 23:42










1 Answer
1






active

oldest

votes


















0












$begingroup$

Thanks for help from @StubbornAtom.

The given $M_X(t)$ is generator of Normal distribution with $mu=2$ and $sigma^2=4$ so this given probability is:
$$P{|X -frac{1}{2}|>2} = 1 - P{|X -frac{1}{2}| le 2} =
1 - P{-frac{3}{2} le X le frac{5}{2}}$$



If we convert $X$ to standard normal distribution with this change:
$$Z=frac{X-mu}{sigma}=frac{X-2}{2} $$
so given probability is:
$$1 - P{-frac{3}{2} le X le frac{5}{2}} = 1 - P{-1.7 le Z le 0.25} =
1-(P{Z le0.25} - P{Z le-1.7}) = 1 - (0.5987 - 0.0446) = 1 - 0.5541 = 0.4459$$



Thanks for help.






share|cite|improve this answer









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    $begingroup$

    Thanks for help from @StubbornAtom.

    The given $M_X(t)$ is generator of Normal distribution with $mu=2$ and $sigma^2=4$ so this given probability is:
    $$P{|X -frac{1}{2}|>2} = 1 - P{|X -frac{1}{2}| le 2} =
    1 - P{-frac{3}{2} le X le frac{5}{2}}$$



    If we convert $X$ to standard normal distribution with this change:
    $$Z=frac{X-mu}{sigma}=frac{X-2}{2} $$
    so given probability is:
    $$1 - P{-frac{3}{2} le X le frac{5}{2}} = 1 - P{-1.7 le Z le 0.25} =
    1-(P{Z le0.25} - P{Z le-1.7}) = 1 - (0.5987 - 0.0446) = 1 - 0.5541 = 0.4459$$



    Thanks for help.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Thanks for help from @StubbornAtom.

      The given $M_X(t)$ is generator of Normal distribution with $mu=2$ and $sigma^2=4$ so this given probability is:
      $$P{|X -frac{1}{2}|>2} = 1 - P{|X -frac{1}{2}| le 2} =
      1 - P{-frac{3}{2} le X le frac{5}{2}}$$



      If we convert $X$ to standard normal distribution with this change:
      $$Z=frac{X-mu}{sigma}=frac{X-2}{2} $$
      so given probability is:
      $$1 - P{-frac{3}{2} le X le frac{5}{2}} = 1 - P{-1.7 le Z le 0.25} =
      1-(P{Z le0.25} - P{Z le-1.7}) = 1 - (0.5987 - 0.0446) = 1 - 0.5541 = 0.4459$$



      Thanks for help.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Thanks for help from @StubbornAtom.

        The given $M_X(t)$ is generator of Normal distribution with $mu=2$ and $sigma^2=4$ so this given probability is:
        $$P{|X -frac{1}{2}|>2} = 1 - P{|X -frac{1}{2}| le 2} =
        1 - P{-frac{3}{2} le X le frac{5}{2}}$$



        If we convert $X$ to standard normal distribution with this change:
        $$Z=frac{X-mu}{sigma}=frac{X-2}{2} $$
        so given probability is:
        $$1 - P{-frac{3}{2} le X le frac{5}{2}} = 1 - P{-1.7 le Z le 0.25} =
        1-(P{Z le0.25} - P{Z le-1.7}) = 1 - (0.5987 - 0.0446) = 1 - 0.5541 = 0.4459$$



        Thanks for help.






        share|cite|improve this answer









        $endgroup$



        Thanks for help from @StubbornAtom.

        The given $M_X(t)$ is generator of Normal distribution with $mu=2$ and $sigma^2=4$ so this given probability is:
        $$P{|X -frac{1}{2}|>2} = 1 - P{|X -frac{1}{2}| le 2} =
        1 - P{-frac{3}{2} le X le frac{5}{2}}$$



        If we convert $X$ to standard normal distribution with this change:
        $$Z=frac{X-mu}{sigma}=frac{X-2}{2} $$
        so given probability is:
        $$1 - P{-frac{3}{2} le X le frac{5}{2}} = 1 - P{-1.7 le Z le 0.25} =
        1-(P{Z le0.25} - P{Z le-1.7}) = 1 - (0.5987 - 0.0446) = 1 - 0.5541 = 0.4459$$



        Thanks for help.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 11:30









        AminAmin

        355110




        355110






























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            Cáceres

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            Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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