Finding the joint mgf of two random variables given conditions












0












$begingroup$



Let $X_{1}$ and $X_{2}$ be two independently and identically
distributed random variables with a common



$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2})$$



for all $x in (-infty, infty)$.
Let $Y_{1} = aX_{1} + bX_{2}$ and $Y_{2} = cX_{1} + dX_{2}$ where parameters $a, b, c, d$ satisfy $a^2 + b^2 = 1$ and $c^2 + d^2 =
1$ and $ac + bd = 0$. Find the joint mgf of $(Y_{1}, Y_{2})$.




I need help solving this problem. So we want to find



$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{s(aX_{1} + bX_{2})} cdot e^{t(cX_{1} + dX_{2})}].$$



I don't think we can separate this into the product of two expectations because that would assume $Y_{1}$ and $Y_{2}$ are independent. Also, we can easily find the joint pdf of $X_{1}$ and $X_{2}$, which is just $f(x)^{2}$. I'm just not sure about how to do this problem, and I can't make use of the conditions.




EDIT: We can write



$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{(sa + tc)X_{1}}] cdot mathbb{E}[e^{(sb + td)X_{2}}].$$



First we compute $mathbb{E}[e^{(sa + tc)X_{1}}]$:



$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{(sa + tc)x_{1} - x_{1}^{2}} mathop{dx_{1}} $$



EDIT 2: Using an integration table, we get



$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{e^{(sa + tc)^{2}}}{sqrt{2}}$$



Similarly,



$$mathbb{E}[e^{(sb + td)X_{2}}] = frac{e^{(sb + td)^{2}}}{sqrt{2}}$$



So, the moment generating function is given by



$$frac{e^{(sb + td)^{2}}}{sqrt{2}} frac{e^{(sa + tc)^{2}}}{sqrt{2}} = frac{e^{s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2}}{2}$$



But, the exponent of $e$ can be rewritten as



$$s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2 = s^{2}(a^{2} + b^{2}) + t^2(c^2 + d^2) + 2st(ac + bd) = s^2 + t^2. $$



So, our moment generating function is given by



$$e^{s^2 + t^2}/2 $$






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
    $endgroup$
    – Tki Deneb
    Dec 16 '18 at 19:50












  • $begingroup$
    @TkiDeneb think i figured it out ur way. can u check?
    $endgroup$
    – user614735
    Dec 16 '18 at 20:18










  • $begingroup$
    I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
    $endgroup$
    – Tki Deneb
    Dec 16 '18 at 20:46












  • $begingroup$
    okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
    $endgroup$
    – user614735
    Dec 16 '18 at 20:51
















0












$begingroup$



Let $X_{1}$ and $X_{2}$ be two independently and identically
distributed random variables with a common



$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2})$$



for all $x in (-infty, infty)$.
Let $Y_{1} = aX_{1} + bX_{2}$ and $Y_{2} = cX_{1} + dX_{2}$ where parameters $a, b, c, d$ satisfy $a^2 + b^2 = 1$ and $c^2 + d^2 =
1$ and $ac + bd = 0$. Find the joint mgf of $(Y_{1}, Y_{2})$.




I need help solving this problem. So we want to find



$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{s(aX_{1} + bX_{2})} cdot e^{t(cX_{1} + dX_{2})}].$$



I don't think we can separate this into the product of two expectations because that would assume $Y_{1}$ and $Y_{2}$ are independent. Also, we can easily find the joint pdf of $X_{1}$ and $X_{2}$, which is just $f(x)^{2}$. I'm just not sure about how to do this problem, and I can't make use of the conditions.




EDIT: We can write



$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{(sa + tc)X_{1}}] cdot mathbb{E}[e^{(sb + td)X_{2}}].$$



First we compute $mathbb{E}[e^{(sa + tc)X_{1}}]$:



$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{(sa + tc)x_{1} - x_{1}^{2}} mathop{dx_{1}} $$



EDIT 2: Using an integration table, we get



$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{e^{(sa + tc)^{2}}}{sqrt{2}}$$



Similarly,



$$mathbb{E}[e^{(sb + td)X_{2}}] = frac{e^{(sb + td)^{2}}}{sqrt{2}}$$



So, the moment generating function is given by



$$frac{e^{(sb + td)^{2}}}{sqrt{2}} frac{e^{(sa + tc)^{2}}}{sqrt{2}} = frac{e^{s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2}}{2}$$



But, the exponent of $e$ can be rewritten as



$$s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2 = s^{2}(a^{2} + b^{2}) + t^2(c^2 + d^2) + 2st(ac + bd) = s^2 + t^2. $$



So, our moment generating function is given by



$$e^{s^2 + t^2}/2 $$






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
    $endgroup$
    – Tki Deneb
    Dec 16 '18 at 19:50












  • $begingroup$
    @TkiDeneb think i figured it out ur way. can u check?
    $endgroup$
    – user614735
    Dec 16 '18 at 20:18










  • $begingroup$
    I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
    $endgroup$
    – Tki Deneb
    Dec 16 '18 at 20:46












  • $begingroup$
    okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
    $endgroup$
    – user614735
    Dec 16 '18 at 20:51














0












0








0





$begingroup$



Let $X_{1}$ and $X_{2}$ be two independently and identically
distributed random variables with a common



$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2})$$



for all $x in (-infty, infty)$.
Let $Y_{1} = aX_{1} + bX_{2}$ and $Y_{2} = cX_{1} + dX_{2}$ where parameters $a, b, c, d$ satisfy $a^2 + b^2 = 1$ and $c^2 + d^2 =
1$ and $ac + bd = 0$. Find the joint mgf of $(Y_{1}, Y_{2})$.




I need help solving this problem. So we want to find



$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{s(aX_{1} + bX_{2})} cdot e^{t(cX_{1} + dX_{2})}].$$



I don't think we can separate this into the product of two expectations because that would assume $Y_{1}$ and $Y_{2}$ are independent. Also, we can easily find the joint pdf of $X_{1}$ and $X_{2}$, which is just $f(x)^{2}$. I'm just not sure about how to do this problem, and I can't make use of the conditions.




EDIT: We can write



$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{(sa + tc)X_{1}}] cdot mathbb{E}[e^{(sb + td)X_{2}}].$$



First we compute $mathbb{E}[e^{(sa + tc)X_{1}}]$:



$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{(sa + tc)x_{1} - x_{1}^{2}} mathop{dx_{1}} $$



EDIT 2: Using an integration table, we get



$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{e^{(sa + tc)^{2}}}{sqrt{2}}$$



Similarly,



$$mathbb{E}[e^{(sb + td)X_{2}}] = frac{e^{(sb + td)^{2}}}{sqrt{2}}$$



So, the moment generating function is given by



$$frac{e^{(sb + td)^{2}}}{sqrt{2}} frac{e^{(sa + tc)^{2}}}{sqrt{2}} = frac{e^{s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2}}{2}$$



But, the exponent of $e$ can be rewritten as



$$s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2 = s^{2}(a^{2} + b^{2}) + t^2(c^2 + d^2) + 2st(ac + bd) = s^2 + t^2. $$



So, our moment generating function is given by



$$e^{s^2 + t^2}/2 $$






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$





Let $X_{1}$ and $X_{2}$ be two independently and identically
distributed random variables with a common



$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2})$$



for all $x in (-infty, infty)$.
Let $Y_{1} = aX_{1} + bX_{2}$ and $Y_{2} = cX_{1} + dX_{2}$ where parameters $a, b, c, d$ satisfy $a^2 + b^2 = 1$ and $c^2 + d^2 =
1$ and $ac + bd = 0$. Find the joint mgf of $(Y_{1}, Y_{2})$.




I need help solving this problem. So we want to find



$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{s(aX_{1} + bX_{2})} cdot e^{t(cX_{1} + dX_{2})}].$$



I don't think we can separate this into the product of two expectations because that would assume $Y_{1}$ and $Y_{2}$ are independent. Also, we can easily find the joint pdf of $X_{1}$ and $X_{2}$, which is just $f(x)^{2}$. I'm just not sure about how to do this problem, and I can't make use of the conditions.




EDIT: We can write



$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{(sa + tc)X_{1}}] cdot mathbb{E}[e^{(sb + td)X_{2}}].$$



First we compute $mathbb{E}[e^{(sa + tc)X_{1}}]$:



$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{(sa + tc)x_{1} - x_{1}^{2}} mathop{dx_{1}} $$



EDIT 2: Using an integration table, we get



$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{e^{(sa + tc)^{2}}}{sqrt{2}}$$



Similarly,



$$mathbb{E}[e^{(sb + td)X_{2}}] = frac{e^{(sb + td)^{2}}}{sqrt{2}}$$



So, the moment generating function is given by



$$frac{e^{(sb + td)^{2}}}{sqrt{2}} frac{e^{(sa + tc)^{2}}}{sqrt{2}} = frac{e^{s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2}}{2}$$



But, the exponent of $e$ can be rewritten as



$$s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2 = s^{2}(a^{2} + b^{2}) + t^2(c^2 + d^2) + 2st(ac + bd) = s^2 + t^2. $$



So, our moment generating function is given by



$$e^{s^2 + t^2}/2 $$






probability probability-theory probability-distributions




probability probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 20:18

























asked Dec 16 '18 at 19:34







user614735



















  • $begingroup$
    Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
    $endgroup$
    – Tki Deneb
    Dec 16 '18 at 19:50












  • $begingroup$
    @TkiDeneb think i figured it out ur way. can u check?
    $endgroup$
    – user614735
    Dec 16 '18 at 20:18










  • $begingroup$
    I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
    $endgroup$
    – Tki Deneb
    Dec 16 '18 at 20:46












  • $begingroup$
    okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
    $endgroup$
    – user614735
    Dec 16 '18 at 20:51


















  • $begingroup$
    Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
    $endgroup$
    – Tki Deneb
    Dec 16 '18 at 19:50












  • $begingroup$
    @TkiDeneb think i figured it out ur way. can u check?
    $endgroup$
    – user614735
    Dec 16 '18 at 20:18










  • $begingroup$
    I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
    $endgroup$
    – Tki Deneb
    Dec 16 '18 at 20:46












  • $begingroup$
    okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
    $endgroup$
    – user614735
    Dec 16 '18 at 20:51
















$begingroup$
Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
$endgroup$
– Tki Deneb
Dec 16 '18 at 19:50






$begingroup$
Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
$endgroup$
– Tki Deneb
Dec 16 '18 at 19:50














$begingroup$
@TkiDeneb think i figured it out ur way. can u check?
$endgroup$
– user614735
Dec 16 '18 at 20:18




$begingroup$
@TkiDeneb think i figured it out ur way. can u check?
$endgroup$
– user614735
Dec 16 '18 at 20:18












$begingroup$
I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
$endgroup$
– Tki Deneb
Dec 16 '18 at 20:46






$begingroup$
I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
$endgroup$
– Tki Deneb
Dec 16 '18 at 20:46














$begingroup$
okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
$endgroup$
– user614735
Dec 16 '18 at 20:51




$begingroup$
okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
$endgroup$
– user614735
Dec 16 '18 at 20:51










1 Answer
1






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oldest

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$begingroup$

We can directly use the indepencence of $X_1$ and $X_2$:



$$
Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
$$



Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
$$
Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
$$






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






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    0












    $begingroup$

    We can directly use the indepencence of $X_1$ and $X_2$:



    $$
    Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
    $$



    Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
    $$
    Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We can directly use the indepencence of $X_1$ and $X_2$:



      $$
      Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
      $$



      Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
      $$
      Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We can directly use the indepencence of $X_1$ and $X_2$:



        $$
        Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
        $$



        Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
        $$
        Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
        $$






        share|cite|improve this answer









        $endgroup$



        We can directly use the indepencence of $X_1$ and $X_2$:



        $$
        Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
        $$



        Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
        $$
        Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 21:10









        Tki DenebTki Deneb

        31710




        31710






























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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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