Proving equation for normal distributed variables












0














I'm struggling at finishing an exercise and hope some of you can help me!



Let be $X_i$ a set of random variables being normal distributed $N(0, sigma_i^2$ for $i = 1,...,n$. I need to prove, that $mathbb{E}[max_{1leq i leq n}|X_i|] leq sqrt{6 log(n)} max_{1leq i leq n}sigma_i $.



I followed a hint and proved for a konvex $psi: mathbb{R}^{+}rightarrow mathbb{R}^{+}$ that, if there exist $c_i > 0, i=1, … ,n$ and $c>0$, such that $mathbb{E}[frac{|X_i|}{c_i}] < c, forall i = 1,...,n $ it holds:



$mathbb{E}[max_{1leq i leq n}|X_i|] leq psi^{-1}(c~ n) max_{1leq i leq n}c_i$



Now I tried to apply this little lemma and took $psi(x) = e^{x^2}$. This one is convex and strictly positive and has the inverse $psi^{-1}(y) = sqrt{log(y)}$.



I guess now I should take into account, that the $X_i$ are normal distributed. Unfortunatelly I didn't get the expected result, so it seems as if I would do something wrong...



Can anyone help me?



Thank you very much! :)






measure-theory random-variables normal-distribution







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    0














    I'm struggling at finishing an exercise and hope some of you can help me!



    Let be $X_i$ a set of random variables being normal distributed $N(0, sigma_i^2$ for $i = 1,...,n$. I need to prove, that $mathbb{E}[max_{1leq i leq n}|X_i|] leq sqrt{6 log(n)} max_{1leq i leq n}sigma_i $.



    I followed a hint and proved for a konvex $psi: mathbb{R}^{+}rightarrow mathbb{R}^{+}$ that, if there exist $c_i > 0, i=1, … ,n$ and $c>0$, such that $mathbb{E}[frac{|X_i|}{c_i}] < c, forall i = 1,...,n $ it holds:



    $mathbb{E}[max_{1leq i leq n}|X_i|] leq psi^{-1}(c~ n) max_{1leq i leq n}c_i$



    Now I tried to apply this little lemma and took $psi(x) = e^{x^2}$. This one is convex and strictly positive and has the inverse $psi^{-1}(y) = sqrt{log(y)}$.



    I guess now I should take into account, that the $X_i$ are normal distributed. Unfortunatelly I didn't get the expected result, so it seems as if I would do something wrong...



    Can anyone help me?



    Thank you very much! :)






    measure-theory random-variables normal-distribution







    share|cite|improve this question

























      0












      0








      0







      I'm struggling at finishing an exercise and hope some of you can help me!



      Let be $X_i$ a set of random variables being normal distributed $N(0, sigma_i^2$ for $i = 1,...,n$. I need to prove, that $mathbb{E}[max_{1leq i leq n}|X_i|] leq sqrt{6 log(n)} max_{1leq i leq n}sigma_i $.



      I followed a hint and proved for a konvex $psi: mathbb{R}^{+}rightarrow mathbb{R}^{+}$ that, if there exist $c_i > 0, i=1, … ,n$ and $c>0$, such that $mathbb{E}[frac{|X_i|}{c_i}] < c, forall i = 1,...,n $ it holds:



      $mathbb{E}[max_{1leq i leq n}|X_i|] leq psi^{-1}(c~ n) max_{1leq i leq n}c_i$



      Now I tried to apply this little lemma and took $psi(x) = e^{x^2}$. This one is convex and strictly positive and has the inverse $psi^{-1}(y) = sqrt{log(y)}$.



      I guess now I should take into account, that the $X_i$ are normal distributed. Unfortunatelly I didn't get the expected result, so it seems as if I would do something wrong...



      Can anyone help me?



      Thank you very much! :)






      measure-theory random-variables normal-distribution







      share|cite|improve this question













      I'm struggling at finishing an exercise and hope some of you can help me!



      Let be $X_i$ a set of random variables being normal distributed $N(0, sigma_i^2$ for $i = 1,...,n$. I need to prove, that $mathbb{E}[max_{1leq i leq n}|X_i|] leq sqrt{6 log(n)} max_{1leq i leq n}sigma_i $.



      I followed a hint and proved for a konvex $psi: mathbb{R}^{+}rightarrow mathbb{R}^{+}$ that, if there exist $c_i > 0, i=1, … ,n$ and $c>0$, such that $mathbb{E}[frac{|X_i|}{c_i}] < c, forall i = 1,...,n $ it holds:



      $mathbb{E}[max_{1leq i leq n}|X_i|] leq psi^{-1}(c~ n) max_{1leq i leq n}c_i$



      Now I tried to apply this little lemma and took $psi(x) = e^{x^2}$. This one is convex and strictly positive and has the inverse $psi^{-1}(y) = sqrt{log(y)}$.



      I guess now I should take into account, that the $X_i$ are normal distributed. Unfortunatelly I didn't get the expected result, so it seems as if I would do something wrong...



      Can anyone help me?



      Thank you very much! :)






      measure-theory random-variables normal-distribution




      measure-theory random-variables normal-distribution






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      asked Nov 26 at 22:54









      pcalc

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          I don't know why you didn't get the expected result. Take $c_i=sigma_i$. Since $E|X_i| leq sqrt {EX_i^{2}}=sigma_i$ the hypothesis of the lemma holds for any $c>1$ so we get $Emax_i |X_i| leq sqrt {log, cn} max_i sigma_i$ for any $c >1$. Letting $c to 1$ we get a stronger result than what we are asked to prove.






          share|cite|improve this answer





















          • Hi! I was some kind of confused, due to I got a much "better" result, than expected, so I thought I made some (stupid?) misstake. Thank you very much for your help!
            – pcalc
            Nov 27 at 17:28











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          I don't know why you didn't get the expected result. Take $c_i=sigma_i$. Since $E|X_i| leq sqrt {EX_i^{2}}=sigma_i$ the hypothesis of the lemma holds for any $c>1$ so we get $Emax_i |X_i| leq sqrt {log, cn} max_i sigma_i$ for any $c >1$. Letting $c to 1$ we get a stronger result than what we are asked to prove.






          share|cite|improve this answer





















          • Hi! I was some kind of confused, due to I got a much "better" result, than expected, so I thought I made some (stupid?) misstake. Thank you very much for your help!
            – pcalc
            Nov 27 at 17:28
















          1














          I don't know why you didn't get the expected result. Take $c_i=sigma_i$. Since $E|X_i| leq sqrt {EX_i^{2}}=sigma_i$ the hypothesis of the lemma holds for any $c>1$ so we get $Emax_i |X_i| leq sqrt {log, cn} max_i sigma_i$ for any $c >1$. Letting $c to 1$ we get a stronger result than what we are asked to prove.






          share|cite|improve this answer





















          • Hi! I was some kind of confused, due to I got a much "better" result, than expected, so I thought I made some (stupid?) misstake. Thank you very much for your help!
            – pcalc
            Nov 27 at 17:28














          1












          1








          1






          I don't know why you didn't get the expected result. Take $c_i=sigma_i$. Since $E|X_i| leq sqrt {EX_i^{2}}=sigma_i$ the hypothesis of the lemma holds for any $c>1$ so we get $Emax_i |X_i| leq sqrt {log, cn} max_i sigma_i$ for any $c >1$. Letting $c to 1$ we get a stronger result than what we are asked to prove.






          share|cite|improve this answer












          I don't know why you didn't get the expected result. Take $c_i=sigma_i$. Since $E|X_i| leq sqrt {EX_i^{2}}=sigma_i$ the hypothesis of the lemma holds for any $c>1$ so we get $Emax_i |X_i| leq sqrt {log, cn} max_i sigma_i$ for any $c >1$. Letting $c to 1$ we get a stronger result than what we are asked to prove.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 23:25









          Kavi Rama Murthy

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          49.6k31854












          • Hi! I was some kind of confused, due to I got a much "better" result, than expected, so I thought I made some (stupid?) misstake. Thank you very much for your help!
            – pcalc
            Nov 27 at 17:28


















          • Hi! I was some kind of confused, due to I got a much "better" result, than expected, so I thought I made some (stupid?) misstake. Thank you very much for your help!
            – pcalc
            Nov 27 at 17:28
















          Hi! I was some kind of confused, due to I got a much "better" result, than expected, so I thought I made some (stupid?) misstake. Thank you very much for your help!
          – pcalc
          Nov 27 at 17:28




          Hi! I was some kind of confused, due to I got a much "better" result, than expected, so I thought I made some (stupid?) misstake. Thank you very much for your help!
          – pcalc
          Nov 27 at 17:28


















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