Marginal distribution of normal random variable with a normal mean

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I have a question about calculation of conditional density of two normal distributions. I have random variables $X|m sim textN(m,sigma^2)$ and $M sim textN(theta, s^2)$, with conditional and marginal densities given by:



$$beginequation beginaligned
f(x|m) &= frac1sigma sqrt2pi cdot exp Big( -frac12 Big( fracx-msigma Big)^2 Big), \[10pt]
f(m) &= frac1s sqrt2pi cdot exp Big( - frac12 Big( fracm-thetas Big)^2 Big).
endaligned endequation$$



I would like to know the marginal distribution of $X$. I have multiplied the above densities to form the joint density, but I cannot successfully integrate the result to get the marginal density of interest. My intuition tells me that this is a normal distribution with different parameters, but I can't prove it.






probability normal-distribution







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  • In second line, $Msim$ should be $msim$. Cannot edit for 1 character.
    – a_statistician
    1 hour ago
















up vote
3
down vote

favorite












I have a question about calculation of conditional density of two normal distributions. I have random variables $X|m sim textN(m,sigma^2)$ and $M sim textN(theta, s^2)$, with conditional and marginal densities given by:



$$beginequation beginaligned
f(x|m) &= frac1sigma sqrt2pi cdot exp Big( -frac12 Big( fracx-msigma Big)^2 Big), \[10pt]
f(m) &= frac1s sqrt2pi cdot exp Big( - frac12 Big( fracm-thetas Big)^2 Big).
endaligned endequation$$



I would like to know the marginal distribution of $X$. I have multiplied the above densities to form the joint density, but I cannot successfully integrate the result to get the marginal density of interest. My intuition tells me that this is a normal distribution with different parameters, but I can't prove it.






probability normal-distribution







share|cite|improve this question









New contributor




Mangnier Loïc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • In second line, $Msim$ should be $msim$. Cannot edit for 1 character.
    – a_statistician
    1 hour ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I have a question about calculation of conditional density of two normal distributions. I have random variables $X|m sim textN(m,sigma^2)$ and $M sim textN(theta, s^2)$, with conditional and marginal densities given by:



$$beginequation beginaligned
f(x|m) &= frac1sigma sqrt2pi cdot exp Big( -frac12 Big( fracx-msigma Big)^2 Big), \[10pt]
f(m) &= frac1s sqrt2pi cdot exp Big( - frac12 Big( fracm-thetas Big)^2 Big).
endaligned endequation$$



I would like to know the marginal distribution of $X$. I have multiplied the above densities to form the joint density, but I cannot successfully integrate the result to get the marginal density of interest. My intuition tells me that this is a normal distribution with different parameters, but I can't prove it.






probability normal-distribution







share|cite|improve this question









New contributor




Mangnier Loïc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I have a question about calculation of conditional density of two normal distributions. I have random variables $X|m sim textN(m,sigma^2)$ and $M sim textN(theta, s^2)$, with conditional and marginal densities given by:



$$beginequation beginaligned
f(x|m) &= frac1sigma sqrt2pi cdot exp Big( -frac12 Big( fracx-msigma Big)^2 Big), \[10pt]
f(m) &= frac1s sqrt2pi cdot exp Big( - frac12 Big( fracm-thetas Big)^2 Big).
endaligned endequation$$



I would like to know the marginal distribution of $X$. I have multiplied the above densities to form the joint density, but I cannot successfully integrate the result to get the marginal density of interest. My intuition tells me that this is a normal distribution with different parameters, but I can't prove it.






probability normal-distribution




probability normal-distribution






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Mangnier Loïc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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Check out our Code of Conduct.









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edited 1 hour ago









Ben

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Check out our Code of Conduct.











  • In second line, $Msim$ should be $msim$. Cannot edit for 1 character.
    – a_statistician
    1 hour ago
















  • In second line, $Msim$ should be $msim$. Cannot edit for 1 character.
    – a_statistician
    1 hour ago















In second line, $Msim$ should be $msim$. Cannot edit for 1 character.
– a_statistician
1 hour ago




In second line, $Msim$ should be $msim$. Cannot edit for 1 character.
– a_statistician
1 hour ago










2 Answers
2






active

oldest

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up vote
1
down vote













Let
$$X = m +epsilon$$
where $m sim N(theta,s^2)$ and $epsilon sim N(0,sigma^2)$ and they are independent.



Then $X|m$ and $m$ follows the distributions specified in the question.



$E(X)=E(m) = theta$



$Var(X) = Var(m) +Var(epsilon) = s^2+sigma^2$



According to "The sum of random variables following Normal distribution follows Normal distribution", and the normal distribution is determined by mean and variance, we have



$$X sim N(theta, s^2+sigma^2)$$






share|cite|improve this answer






















  • Ok thanks, I understand your answer but I would like to have a mathematical proof fo your first line X=m+ϵ. Is it possible ?
    – Mangnier Loïc
    2 hours ago










  • In math prove, some time to need to create some thing. For example, + x - x, divides by a then times a.... So here we need to construct a X, such that 1) the original properties are kept and 2) give us the convenience to proving. Do you mean to prove "Then $X|m$ and $m$ follows the distributions specified in the question."?
    – a_statistician
    2 hours ago

















up vote
1
down vote













Your intuition is correct - the marginal distribution of a normal random variable with a normal mean is indeed normal. To see this, we first re-frame the joint distribution as a product of normal densities by completing the square:



$$beginequation beginaligned
f(x,m)
&= f(x|m) f(m) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( fracx-msigma Big)^2 + Big( fracm-thetas Big)^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( frac1sigma^2+frac1s^2 Big) m^2 -2 Big( fracxsigma^2 + fracthetas^2 Big) m + Big( fracx^2sigma^2 + fractheta^2s^2 Big) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 sigma^2 s^2 Big[ (s^2+sigma^2) m^2 -2 (x s^2+ theta sigma^2) m + (x^2 s^2+ theta^2 sigma^2) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big[ m^2 -2 cdot fracx s^2 + theta sigma^2s^2+sigma^2 cdot m + fracx^2 s^2 + theta^2 sigma^2s^2+sigma^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad quad quad text times exp Big( frac(x s^2 + theta sigma^2)^22 sigma^2 s^2 (s^2+theta^2) - fracx^2 s^2 + theta^2 sigma^22 sigma^2 s^2 Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= sqrtfracs^2+sigma^22pi sigma^2 s^2 cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad times sqrtfrac12pi (s^2+sigma^2) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= textN Big( m Big| fracxs^2+thetasigma^2s^2+sigma^2, fracs^2 sigma^2s^2+sigma^2 Big) cdot textN(x|theta, s^2+sigma^2).
endaligned endequation$$



We then integrate out $m$ to obtain the marginal density $f(x) = textN(x|theta, s^2+sigma^2)$. From this exercise we see that $X sim textN(theta, s^2+sigma^2)$.






share|cite|improve this answer




















  • (+1), would this have been simpler using the moment generating function (MGF)?
    – SecretAgentMan
    27 mins ago










  • @SecretAgentMan: Thanks for the up-votes. The OP said he was having trouble integrating the joint density, so that is the method I have used.
    – Ben
    25 mins ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













Let
$$X = m +epsilon$$
where $m sim N(theta,s^2)$ and $epsilon sim N(0,sigma^2)$ and they are independent.



Then $X|m$ and $m$ follows the distributions specified in the question.



$E(X)=E(m) = theta$



$Var(X) = Var(m) +Var(epsilon) = s^2+sigma^2$



According to "The sum of random variables following Normal distribution follows Normal distribution", and the normal distribution is determined by mean and variance, we have



$$X sim N(theta, s^2+sigma^2)$$






share|cite|improve this answer






















  • Ok thanks, I understand your answer but I would like to have a mathematical proof fo your first line X=m+ϵ. Is it possible ?
    – Mangnier Loïc
    2 hours ago










  • In math prove, some time to need to create some thing. For example, + x - x, divides by a then times a.... So here we need to construct a X, such that 1) the original properties are kept and 2) give us the convenience to proving. Do you mean to prove "Then $X|m$ and $m$ follows the distributions specified in the question."?
    – a_statistician
    2 hours ago














up vote
1
down vote













Let
$$X = m +epsilon$$
where $m sim N(theta,s^2)$ and $epsilon sim N(0,sigma^2)$ and they are independent.



Then $X|m$ and $m$ follows the distributions specified in the question.



$E(X)=E(m) = theta$



$Var(X) = Var(m) +Var(epsilon) = s^2+sigma^2$



According to "The sum of random variables following Normal distribution follows Normal distribution", and the normal distribution is determined by mean and variance, we have



$$X sim N(theta, s^2+sigma^2)$$






share|cite|improve this answer






















  • Ok thanks, I understand your answer but I would like to have a mathematical proof fo your first line X=m+ϵ. Is it possible ?
    – Mangnier Loïc
    2 hours ago










  • In math prove, some time to need to create some thing. For example, + x - x, divides by a then times a.... So here we need to construct a X, such that 1) the original properties are kept and 2) give us the convenience to proving. Do you mean to prove "Then $X|m$ and $m$ follows the distributions specified in the question."?
    – a_statistician
    2 hours ago












up vote
1
down vote










up vote
1
down vote









Let
$$X = m +epsilon$$
where $m sim N(theta,s^2)$ and $epsilon sim N(0,sigma^2)$ and they are independent.



Then $X|m$ and $m$ follows the distributions specified in the question.



$E(X)=E(m) = theta$



$Var(X) = Var(m) +Var(epsilon) = s^2+sigma^2$



According to "The sum of random variables following Normal distribution follows Normal distribution", and the normal distribution is determined by mean and variance, we have



$$X sim N(theta, s^2+sigma^2)$$






share|cite|improve this answer














Let
$$X = m +epsilon$$
where $m sim N(theta,s^2)$ and $epsilon sim N(0,sigma^2)$ and they are independent.



Then $X|m$ and $m$ follows the distributions specified in the question.



$E(X)=E(m) = theta$



$Var(X) = Var(m) +Var(epsilon) = s^2+sigma^2$



According to "The sum of random variables following Normal distribution follows Normal distribution", and the normal distribution is determined by mean and variance, we have



$$X sim N(theta, s^2+sigma^2)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 3 hours ago









a_statistician

1,752139




1,752139











  • Ok thanks, I understand your answer but I would like to have a mathematical proof fo your first line X=m+ϵ. Is it possible ?
    – Mangnier Loïc
    2 hours ago










  • In math prove, some time to need to create some thing. For example, + x - x, divides by a then times a.... So here we need to construct a X, such that 1) the original properties are kept and 2) give us the convenience to proving. Do you mean to prove "Then $X|m$ and $m$ follows the distributions specified in the question."?
    – a_statistician
    2 hours ago
















  • Ok thanks, I understand your answer but I would like to have a mathematical proof fo your first line X=m+ϵ. Is it possible ?
    – Mangnier Loïc
    2 hours ago










  • In math prove, some time to need to create some thing. For example, + x - x, divides by a then times a.... So here we need to construct a X, such that 1) the original properties are kept and 2) give us the convenience to proving. Do you mean to prove "Then $X|m$ and $m$ follows the distributions specified in the question."?
    – a_statistician
    2 hours ago















Ok thanks, I understand your answer but I would like to have a mathematical proof fo your first line X=m+ϵ. Is it possible ?
– Mangnier Loïc
2 hours ago




Ok thanks, I understand your answer but I would like to have a mathematical proof fo your first line X=m+ϵ. Is it possible ?
– Mangnier Loïc
2 hours ago












In math prove, some time to need to create some thing. For example, + x - x, divides by a then times a.... So here we need to construct a X, such that 1) the original properties are kept and 2) give us the convenience to proving. Do you mean to prove "Then $X|m$ and $m$ follows the distributions specified in the question."?
– a_statistician
2 hours ago




In math prove, some time to need to create some thing. For example, + x - x, divides by a then times a.... So here we need to construct a X, such that 1) the original properties are kept and 2) give us the convenience to proving. Do you mean to prove "Then $X|m$ and $m$ follows the distributions specified in the question."?
– a_statistician
2 hours ago












up vote
1
down vote













Your intuition is correct - the marginal distribution of a normal random variable with a normal mean is indeed normal. To see this, we first re-frame the joint distribution as a product of normal densities by completing the square:



$$beginequation beginaligned
f(x,m)
&= f(x|m) f(m) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( fracx-msigma Big)^2 + Big( fracm-thetas Big)^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( frac1sigma^2+frac1s^2 Big) m^2 -2 Big( fracxsigma^2 + fracthetas^2 Big) m + Big( fracx^2sigma^2 + fractheta^2s^2 Big) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 sigma^2 s^2 Big[ (s^2+sigma^2) m^2 -2 (x s^2+ theta sigma^2) m + (x^2 s^2+ theta^2 sigma^2) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big[ m^2 -2 cdot fracx s^2 + theta sigma^2s^2+sigma^2 cdot m + fracx^2 s^2 + theta^2 sigma^2s^2+sigma^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad quad quad text times exp Big( frac(x s^2 + theta sigma^2)^22 sigma^2 s^2 (s^2+theta^2) - fracx^2 s^2 + theta^2 sigma^22 sigma^2 s^2 Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= sqrtfracs^2+sigma^22pi sigma^2 s^2 cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad times sqrtfrac12pi (s^2+sigma^2) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= textN Big( m Big| fracxs^2+thetasigma^2s^2+sigma^2, fracs^2 sigma^2s^2+sigma^2 Big) cdot textN(x|theta, s^2+sigma^2).
endaligned endequation$$



We then integrate out $m$ to obtain the marginal density $f(x) = textN(x|theta, s^2+sigma^2)$. From this exercise we see that $X sim textN(theta, s^2+sigma^2)$.






share|cite|improve this answer




















  • (+1), would this have been simpler using the moment generating function (MGF)?
    – SecretAgentMan
    27 mins ago










  • @SecretAgentMan: Thanks for the up-votes. The OP said he was having trouble integrating the joint density, so that is the method I have used.
    – Ben
    25 mins ago














up vote
1
down vote













Your intuition is correct - the marginal distribution of a normal random variable with a normal mean is indeed normal. To see this, we first re-frame the joint distribution as a product of normal densities by completing the square:



$$beginequation beginaligned
f(x,m)
&= f(x|m) f(m) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( fracx-msigma Big)^2 + Big( fracm-thetas Big)^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( frac1sigma^2+frac1s^2 Big) m^2 -2 Big( fracxsigma^2 + fracthetas^2 Big) m + Big( fracx^2sigma^2 + fractheta^2s^2 Big) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 sigma^2 s^2 Big[ (s^2+sigma^2) m^2 -2 (x s^2+ theta sigma^2) m + (x^2 s^2+ theta^2 sigma^2) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big[ m^2 -2 cdot fracx s^2 + theta sigma^2s^2+sigma^2 cdot m + fracx^2 s^2 + theta^2 sigma^2s^2+sigma^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad quad quad text times exp Big( frac(x s^2 + theta sigma^2)^22 sigma^2 s^2 (s^2+theta^2) - fracx^2 s^2 + theta^2 sigma^22 sigma^2 s^2 Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= sqrtfracs^2+sigma^22pi sigma^2 s^2 cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad times sqrtfrac12pi (s^2+sigma^2) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= textN Big( m Big| fracxs^2+thetasigma^2s^2+sigma^2, fracs^2 sigma^2s^2+sigma^2 Big) cdot textN(x|theta, s^2+sigma^2).
endaligned endequation$$



We then integrate out $m$ to obtain the marginal density $f(x) = textN(x|theta, s^2+sigma^2)$. From this exercise we see that $X sim textN(theta, s^2+sigma^2)$.






share|cite|improve this answer




















  • (+1), would this have been simpler using the moment generating function (MGF)?
    – SecretAgentMan
    27 mins ago










  • @SecretAgentMan: Thanks for the up-votes. The OP said he was having trouble integrating the joint density, so that is the method I have used.
    – Ben
    25 mins ago












up vote
1
down vote










up vote
1
down vote









Your intuition is correct - the marginal distribution of a normal random variable with a normal mean is indeed normal. To see this, we first re-frame the joint distribution as a product of normal densities by completing the square:



$$beginequation beginaligned
f(x,m)
&= f(x|m) f(m) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( fracx-msigma Big)^2 + Big( fracm-thetas Big)^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( frac1sigma^2+frac1s^2 Big) m^2 -2 Big( fracxsigma^2 + fracthetas^2 Big) m + Big( fracx^2sigma^2 + fractheta^2s^2 Big) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 sigma^2 s^2 Big[ (s^2+sigma^2) m^2 -2 (x s^2+ theta sigma^2) m + (x^2 s^2+ theta^2 sigma^2) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big[ m^2 -2 cdot fracx s^2 + theta sigma^2s^2+sigma^2 cdot m + fracx^2 s^2 + theta^2 sigma^2s^2+sigma^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad quad quad text times exp Big( frac(x s^2 + theta sigma^2)^22 sigma^2 s^2 (s^2+theta^2) - fracx^2 s^2 + theta^2 sigma^22 sigma^2 s^2 Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= sqrtfracs^2+sigma^22pi sigma^2 s^2 cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad times sqrtfrac12pi (s^2+sigma^2) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= textN Big( m Big| fracxs^2+thetasigma^2s^2+sigma^2, fracs^2 sigma^2s^2+sigma^2 Big) cdot textN(x|theta, s^2+sigma^2).
endaligned endequation$$



We then integrate out $m$ to obtain the marginal density $f(x) = textN(x|theta, s^2+sigma^2)$. From this exercise we see that $X sim textN(theta, s^2+sigma^2)$.






share|cite|improve this answer












Your intuition is correct - the marginal distribution of a normal random variable with a normal mean is indeed normal. To see this, we first re-frame the joint distribution as a product of normal densities by completing the square:



$$beginequation beginaligned
f(x,m)
&= f(x|m) f(m) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( fracx-msigma Big)^2 + Big( fracm-thetas Big)^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 Big[ Big( frac1sigma^2+frac1s^2 Big) m^2 -2 Big( fracxsigma^2 + fracthetas^2 Big) m + Big( fracx^2sigma^2 + fractheta^2s^2 Big) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( -frac12 sigma^2 s^2 Big[ (s^2+sigma^2) m^2 -2 (x s^2+ theta sigma^2) m + (x^2 s^2+ theta^2 sigma^2) Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big[ m^2 -2 cdot fracx s^2 + theta sigma^2s^2+sigma^2 cdot m + fracx^2 s^2 + theta^2 sigma^2s^2+sigma^2 Big] Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad quad quad text times exp Big( frac(x s^2 + theta sigma^2)^22 sigma^2 s^2 (s^2+theta^2) - fracx^2 s^2 + theta^2 sigma^22 sigma^2 s^2 Big) \[10pt]
&= frac12pi sigma s cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= sqrtfracs^2+sigma^22pi sigma^2 s^2 cdot exp Big( - fracs^2+sigma^22 sigma^2 s^2 Big( m - fracx s^2 + theta sigma^2s^2+sigma^2 Big)^2 Big) \[6pt]
&quad times sqrtfrac12pi (s^2+sigma^2) cdot exp Big( -frac12 frac(x-theta)^2s^2+sigma^2 Big) \[10pt]
&= textN Big( m Big| fracxs^2+thetasigma^2s^2+sigma^2, fracs^2 sigma^2s^2+sigma^2 Big) cdot textN(x|theta, s^2+sigma^2).
endaligned endequation$$



We then integrate out $m$ to obtain the marginal density $f(x) = textN(x|theta, s^2+sigma^2)$. From this exercise we see that $X sim textN(theta, s^2+sigma^2)$.







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answered 28 mins ago









Ben

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  • (+1), would this have been simpler using the moment generating function (MGF)?
    – SecretAgentMan
    27 mins ago










  • @SecretAgentMan: Thanks for the up-votes. The OP said he was having trouble integrating the joint density, so that is the method I have used.
    – Ben
    25 mins ago
















  • (+1), would this have been simpler using the moment generating function (MGF)?
    – SecretAgentMan
    27 mins ago










  • @SecretAgentMan: Thanks for the up-votes. The OP said he was having trouble integrating the joint density, so that is the method I have used.
    – Ben
    25 mins ago















(+1), would this have been simpler using the moment generating function (MGF)?
– SecretAgentMan
27 mins ago




(+1), would this have been simpler using the moment generating function (MGF)?
– SecretAgentMan
27 mins ago












@SecretAgentMan: Thanks for the up-votes. The OP said he was having trouble integrating the joint density, so that is the method I have used.
– Ben
25 mins ago




@SecretAgentMan: Thanks for the up-votes. The OP said he was having trouble integrating the joint density, so that is the method I have used.
– Ben
25 mins ago










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