$f(x)=alpha e^{-x^2-beta x}$ Find $alpha$ and $beta$ . Expectation is given.












0












$begingroup$



Let the probability density function of a random variable $X$ be given
by



$f(x)=alpha e^{-x^2-beta x} infty<x<infty $



If $E(X)= -dfrac{1}{2}$ then



$(A) alpha =dfrac{e^{frac{-1}{4}}}{sqrt pi}$ and $beta=1$



$(B) alpha =dfrac{e^{frac{-1}{4}}}{sqrt pi}$ and $beta=-1$



$(C) alpha =e^frac{-1}{4}sqrt pi$ and $beta=-1$



$(D) alpha =e^frac{-1}{4}sqrt pi$ and $beta=1$




The way I did this question:



$int_{-infty}^{infty}f(x)dx=int_{-infty}^{infty}alpha e^{-(x^2+beta x +frac{beta^2}{4}-frac{beta^2}{4})}dx=int_{-infty}^{infty}alpha e^{-(x+frac{beta}{2})^2}e^{frac{beta^2}{4}}dx=1 $



put $y=(x+frac{beta}{2})$



$int_{-infty}^{infty}alpha e^{-(y)^2}e^{frac{beta^2}{4}}dx=alphasqrt{pi}e^{frac{beta^2}{4}}=1implies alpha=dfrac{e^{frac{-beta^2}{4}}}{sqrt{pi}}$



Now using $E(X)=int_{-infty}^{infty}xalpha e^{-(x+frac{beta}{2})^2}e^{frac{beta^2}{4}}dx$



put $y=(x+frac{beta}{2})$



$E(X)=int_{-infty}^{infty}(y-frac{beta}{2})alpha e^{-(y)^2}e^{frac{beta^2}{4}}dx=underbrace{alpha e^{frac{beta^2}{4}} int_{-infty}^{infty}(y) e^{-(y)^2}dx}_{text {odd function}}-frac{beta}{2}underbrace{int_{-infty}^{infty}alpha e^{-(y)^2}e^{frac{-beta^2}{4}}}_{text{pdf}}dx$



$0-dfrac{beta}{2}=-dfrac{1}{2} implies fbox {$beta$ =1}$ and $fbox{$alpha=dfrac{e^{frac{-1}{4}}}{sqrt{pi}}$} $



Somehow after doing rigorous calculations, I got this answer but this question came in 2 marks so I am looking for a quick way out. Any alternate solution which less time consuming ?






probability statistics expected-value







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$endgroup$

















    0












    $begingroup$



    Let the probability density function of a random variable $X$ be given
    by



    $f(x)=alpha e^{-x^2-beta x} infty<x<infty $



    If $E(X)= -dfrac{1}{2}$ then



    $(A) alpha =dfrac{e^{frac{-1}{4}}}{sqrt pi}$ and $beta=1$



    $(B) alpha =dfrac{e^{frac{-1}{4}}}{sqrt pi}$ and $beta=-1$



    $(C) alpha =e^frac{-1}{4}sqrt pi$ and $beta=-1$



    $(D) alpha =e^frac{-1}{4}sqrt pi$ and $beta=1$




    The way I did this question:



    $int_{-infty}^{infty}f(x)dx=int_{-infty}^{infty}alpha e^{-(x^2+beta x +frac{beta^2}{4}-frac{beta^2}{4})}dx=int_{-infty}^{infty}alpha e^{-(x+frac{beta}{2})^2}e^{frac{beta^2}{4}}dx=1 $



    put $y=(x+frac{beta}{2})$



    $int_{-infty}^{infty}alpha e^{-(y)^2}e^{frac{beta^2}{4}}dx=alphasqrt{pi}e^{frac{beta^2}{4}}=1implies alpha=dfrac{e^{frac{-beta^2}{4}}}{sqrt{pi}}$



    Now using $E(X)=int_{-infty}^{infty}xalpha e^{-(x+frac{beta}{2})^2}e^{frac{beta^2}{4}}dx$



    put $y=(x+frac{beta}{2})$



    $E(X)=int_{-infty}^{infty}(y-frac{beta}{2})alpha e^{-(y)^2}e^{frac{beta^2}{4}}dx=underbrace{alpha e^{frac{beta^2}{4}} int_{-infty}^{infty}(y) e^{-(y)^2}dx}_{text {odd function}}-frac{beta}{2}underbrace{int_{-infty}^{infty}alpha e^{-(y)^2}e^{frac{-beta^2}{4}}}_{text{pdf}}dx$



    $0-dfrac{beta}{2}=-dfrac{1}{2} implies fbox {$beta$ =1}$ and $fbox{$alpha=dfrac{e^{frac{-1}{4}}}{sqrt{pi}}$} $



    Somehow after doing rigorous calculations, I got this answer but this question came in 2 marks so I am looking for a quick way out. Any alternate solution which less time consuming ?






    probability statistics expected-value







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Let the probability density function of a random variable $X$ be given
      by



      $f(x)=alpha e^{-x^2-beta x} infty<x<infty $



      If $E(X)= -dfrac{1}{2}$ then



      $(A) alpha =dfrac{e^{frac{-1}{4}}}{sqrt pi}$ and $beta=1$



      $(B) alpha =dfrac{e^{frac{-1}{4}}}{sqrt pi}$ and $beta=-1$



      $(C) alpha =e^frac{-1}{4}sqrt pi$ and $beta=-1$



      $(D) alpha =e^frac{-1}{4}sqrt pi$ and $beta=1$




      The way I did this question:



      $int_{-infty}^{infty}f(x)dx=int_{-infty}^{infty}alpha e^{-(x^2+beta x +frac{beta^2}{4}-frac{beta^2}{4})}dx=int_{-infty}^{infty}alpha e^{-(x+frac{beta}{2})^2}e^{frac{beta^2}{4}}dx=1 $



      put $y=(x+frac{beta}{2})$



      $int_{-infty}^{infty}alpha e^{-(y)^2}e^{frac{beta^2}{4}}dx=alphasqrt{pi}e^{frac{beta^2}{4}}=1implies alpha=dfrac{e^{frac{-beta^2}{4}}}{sqrt{pi}}$



      Now using $E(X)=int_{-infty}^{infty}xalpha e^{-(x+frac{beta}{2})^2}e^{frac{beta^2}{4}}dx$



      put $y=(x+frac{beta}{2})$



      $E(X)=int_{-infty}^{infty}(y-frac{beta}{2})alpha e^{-(y)^2}e^{frac{beta^2}{4}}dx=underbrace{alpha e^{frac{beta^2}{4}} int_{-infty}^{infty}(y) e^{-(y)^2}dx}_{text {odd function}}-frac{beta}{2}underbrace{int_{-infty}^{infty}alpha e^{-(y)^2}e^{frac{-beta^2}{4}}}_{text{pdf}}dx$



      $0-dfrac{beta}{2}=-dfrac{1}{2} implies fbox {$beta$ =1}$ and $fbox{$alpha=dfrac{e^{frac{-1}{4}}}{sqrt{pi}}$} $



      Somehow after doing rigorous calculations, I got this answer but this question came in 2 marks so I am looking for a quick way out. Any alternate solution which less time consuming ?






      probability statistics expected-value







      share|cite|improve this question









      $endgroup$





      Let the probability density function of a random variable $X$ be given
      by



      $f(x)=alpha e^{-x^2-beta x} infty<x<infty $



      If $E(X)= -dfrac{1}{2}$ then



      $(A) alpha =dfrac{e^{frac{-1}{4}}}{sqrt pi}$ and $beta=1$



      $(B) alpha =dfrac{e^{frac{-1}{4}}}{sqrt pi}$ and $beta=-1$



      $(C) alpha =e^frac{-1}{4}sqrt pi$ and $beta=-1$



      $(D) alpha =e^frac{-1}{4}sqrt pi$ and $beta=1$




      The way I did this question:



      $int_{-infty}^{infty}f(x)dx=int_{-infty}^{infty}alpha e^{-(x^2+beta x +frac{beta^2}{4}-frac{beta^2}{4})}dx=int_{-infty}^{infty}alpha e^{-(x+frac{beta}{2})^2}e^{frac{beta^2}{4}}dx=1 $



      put $y=(x+frac{beta}{2})$



      $int_{-infty}^{infty}alpha e^{-(y)^2}e^{frac{beta^2}{4}}dx=alphasqrt{pi}e^{frac{beta^2}{4}}=1implies alpha=dfrac{e^{frac{-beta^2}{4}}}{sqrt{pi}}$



      Now using $E(X)=int_{-infty}^{infty}xalpha e^{-(x+frac{beta}{2})^2}e^{frac{beta^2}{4}}dx$



      put $y=(x+frac{beta}{2})$



      $E(X)=int_{-infty}^{infty}(y-frac{beta}{2})alpha e^{-(y)^2}e^{frac{beta^2}{4}}dx=underbrace{alpha e^{frac{beta^2}{4}} int_{-infty}^{infty}(y) e^{-(y)^2}dx}_{text {odd function}}-frac{beta}{2}underbrace{int_{-infty}^{infty}alpha e^{-(y)^2}e^{frac{-beta^2}{4}}}_{text{pdf}}dx$



      $0-dfrac{beta}{2}=-dfrac{1}{2} implies fbox {$beta$ =1}$ and $fbox{$alpha=dfrac{e^{frac{-1}{4}}}{sqrt{pi}}$} $



      Somehow after doing rigorous calculations, I got this answer but this question came in 2 marks so I am looking for a quick way out. Any alternate solution which less time consuming ?






      probability statistics expected-value




      probability statistics expected-value






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 27 at 12:44









      Daman deepDaman deep

      756419




      756419






















          1 Answer
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          $begingroup$

          $f(x)$ is in the form of a normal PDF: $frac{1}{sqrt{2pisigma^2}}exp(-(x-mu)^2/2sigma^2)$, which should be equal to $alphaexp(-x^2-beta x)$. Coefficients should match in LHS and RHS, so $2sigma^2=1$ and $E[X]=mu=-1/2$. Then, LHS becomes
          $$frac{1}{sqrt{pi}}exp(-x^2+2mu x - mu^2)=alphaexp(-x^2-beta x)$$
          which gives you $beta=1$ and $alpha=frac{exp(-1/4)}{sqrt{pi}}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Omg I feel so dumb now .
            $endgroup$
            – Daman deep
            Jan 27 at 13:07










          • $begingroup$
            no you're path independent :)
            $endgroup$
            – gunes
            Jan 27 at 13:08











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          1 Answer
          1






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          active

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          2












          $begingroup$

          $f(x)$ is in the form of a normal PDF: $frac{1}{sqrt{2pisigma^2}}exp(-(x-mu)^2/2sigma^2)$, which should be equal to $alphaexp(-x^2-beta x)$. Coefficients should match in LHS and RHS, so $2sigma^2=1$ and $E[X]=mu=-1/2$. Then, LHS becomes
          $$frac{1}{sqrt{pi}}exp(-x^2+2mu x - mu^2)=alphaexp(-x^2-beta x)$$
          which gives you $beta=1$ and $alpha=frac{exp(-1/4)}{sqrt{pi}}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Omg I feel so dumb now .
            $endgroup$
            – Daman deep
            Jan 27 at 13:07










          • $begingroup$
            no you're path independent :)
            $endgroup$
            – gunes
            Jan 27 at 13:08
















          2












          $begingroup$

          $f(x)$ is in the form of a normal PDF: $frac{1}{sqrt{2pisigma^2}}exp(-(x-mu)^2/2sigma^2)$, which should be equal to $alphaexp(-x^2-beta x)$. Coefficients should match in LHS and RHS, so $2sigma^2=1$ and $E[X]=mu=-1/2$. Then, LHS becomes
          $$frac{1}{sqrt{pi}}exp(-x^2+2mu x - mu^2)=alphaexp(-x^2-beta x)$$
          which gives you $beta=1$ and $alpha=frac{exp(-1/4)}{sqrt{pi}}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Omg I feel so dumb now .
            $endgroup$
            – Daman deep
            Jan 27 at 13:07










          • $begingroup$
            no you're path independent :)
            $endgroup$
            – gunes
            Jan 27 at 13:08














          2












          2








          2





          $begingroup$

          $f(x)$ is in the form of a normal PDF: $frac{1}{sqrt{2pisigma^2}}exp(-(x-mu)^2/2sigma^2)$, which should be equal to $alphaexp(-x^2-beta x)$. Coefficients should match in LHS and RHS, so $2sigma^2=1$ and $E[X]=mu=-1/2$. Then, LHS becomes
          $$frac{1}{sqrt{pi}}exp(-x^2+2mu x - mu^2)=alphaexp(-x^2-beta x)$$
          which gives you $beta=1$ and $alpha=frac{exp(-1/4)}{sqrt{pi}}$.






          share|cite|improve this answer









          $endgroup$



          $f(x)$ is in the form of a normal PDF: $frac{1}{sqrt{2pisigma^2}}exp(-(x-mu)^2/2sigma^2)$, which should be equal to $alphaexp(-x^2-beta x)$. Coefficients should match in LHS and RHS, so $2sigma^2=1$ and $E[X]=mu=-1/2$. Then, LHS becomes
          $$frac{1}{sqrt{pi}}exp(-x^2+2mu x - mu^2)=alphaexp(-x^2-beta x)$$
          which gives you $beta=1$ and $alpha=frac{exp(-1/4)}{sqrt{pi}}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 27 at 13:03









          gunesgunes

          3747




          3747












          • $begingroup$
            Omg I feel so dumb now .
            $endgroup$
            – Daman deep
            Jan 27 at 13:07










          • $begingroup$
            no you're path independent :)
            $endgroup$
            – gunes
            Jan 27 at 13:08


















          • $begingroup$
            Omg I feel so dumb now .
            $endgroup$
            – Daman deep
            Jan 27 at 13:07










          • $begingroup$
            no you're path independent :)
            $endgroup$
            – gunes
            Jan 27 at 13:08
















          $begingroup$
          Omg I feel so dumb now .
          $endgroup$
          – Daman deep
          Jan 27 at 13:07




          $begingroup$
          Omg I feel so dumb now .
          $endgroup$
          – Daman deep
          Jan 27 at 13:07












          $begingroup$
          no you're path independent :)
          $endgroup$
          – gunes
          Jan 27 at 13:08




          $begingroup$
          no you're path independent :)
          $endgroup$
          – gunes
          Jan 27 at 13:08


















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