Product of Two Normal Distributions

**vioravis** · May 12th 2008, 10:53 PM

Hello,

I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.

**mr fantastic** · May 13th 2008, 05:34 AM

Originally Posted by vioravis

Hello,

I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.

No.

If the two random variables X and Y are independent, then the pdf of Z = XY is probably (I haven't done the calculation) a Bessel function. See 3. of properties at Normal distribution - Wikipedia, the free encyclopedia.

**CaptainBlack** · May 13th 2008, 06:13 AM

Originally Posted by vioravis

Hello,

I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.

Do you mean "What is the distribution of the product of two normally distributed RV with means and variances which may be different?"?

RonL

**vioravis** · May 13th 2008, 07:01 AM

I have the following two densities:

f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}

So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).

Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.

**vioravis** · May 13th 2008, 08:01 AM

In Page 3 of the following link, it is given that the product of two gaussian densities is also gaussian. I think it is same as the one I requested above. I am looking for the derivation of the formula given for the resultant distribution:

http://www.cs.cmu.edu/afs/cs/academi...ww/hw5/hw5.pdf

confirmed here as well:

Product of Two Gaussian PDFs

**mr fantastic** · May 13th 2008, 03:54 PM

Originally Posted by vioravis

I have the following two densities:

f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}

So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).

Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.

If I understand you correctly, what you're asking boils down to wanting to show that

$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2} = -C \frac{(x - \mu)^2}{2\sigma^2}$

and getting the appropriate expressions for $\mu, ~ \sigma$ and C in terms of $\mu_1, ~ \mu_2, ~ \sigma_1$ and $\sigma_2$ . Note that $e^C$ becomes part of the normalising constant.

It's simple to show and get the expressions but tedious to type out.

**vioravis** · May 13th 2008, 04:20 PM

Fantastic,

Thanks a lot. Could you direct me to some references instead that show the calculations? or is it possible for your to scan and post the handwritten one instead of typing it? Thank you.

**mr fantastic** · May 14th 2008, 04:42 AM

Originally Posted by mr fantastic

If I understand you correctly, what you're asking boils down to wanting to show that

$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2} = -C \frac{(x - \mu)^2}{2\sigma^2}$

and getting the appropriate expressions for $\mu, ~ \sigma$ and C in terms of $\mu_1, ~ \mu_2, ~ \sigma_1$ and $\sigma_2$ . Note that $e^C$ becomes part of the normalising constant.

It's simple to show and get the expressions but tedious to type out.

$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2}$

$= \frac{-\sigma^2_2(x - \mu_1)^2 - \sigma^2_1 (x - \mu_2)^2}{2 \sigma_1^2 \sigma^2}$

$= \frac{ -\sigma^2_2 x^2 + 2\sigma_2^2 \mu_1 x - \mu_1^2 \sigma_2^2 -\sigma^2_1 x^2 + 2\sigma_1^2 \mu_2 x - \mu_2^2 \sigma_1^2}{2 \sigma_1^2 \sigma^2}$

$= \frac{ -(\sigma_1^2 + \sigma_2^2) x^2 + 2(\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2) x - (\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2)}{2 \sigma_1^2 \sigma^2}$

$= \frac{-(\sigma_1^2 + \sigma_2^2) \left[ x^2 - \frac{2(\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2) x}{\sigma_1^2 + \sigma_2^2} + \frac{\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2}{\sigma_1^2 + \sigma_2^2}\right]}{2 \sigma_1^2 \sigma^2}$

$= \frac{-(\sigma_1^2 + \sigma_2^2) \left[ \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2 - \left(\frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2}{\sigma_1^2 + \sigma_2^2} \right)^2 \right]}{2 \sigma_1^2 \sigma^2}$

$= \frac{- \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2 + \left(\frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2}{\sigma_1^2 + \sigma_2^2} \right)^2}{\frac{2 \sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}}$

$= \frac{- \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2}{\frac{2 \sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}} + C$ .

So $\mu = \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}$ and $\sigma^2 = \frac{\sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}$ .

**CaptainBlack** · May 14th 2008, 05:33 AM

Originally Posted by vioravis

I have the following two densities:

f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}

So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).

Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.

That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.

So we can bust a gut attempting to prove that $\int_{-\infty}^{\infty} f(x)~dx \ne 1$ , but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $1$ so $f$ is not a density.

Code:

>s1=1,mu1=0,s2=2, mu2=5
            1 
            0 
            2 
            5 
>dx=0.2;
>x=-10+dx/2:dx:20;
>
>f1=1/(s1*sqrt(2*pi))* exp( -(x-mu1)^2 / (2*s1^2) );
>f2=1/(s2*sqrt(2*pi))* exp( -(x-mu2)^2/  (2*s2^2) );
>
>f=f1*f2;
>
>II1=sum(f1)*dx
            1 
>II2=sum(f2)*dx
            1 
>II=sum(f)*dx
     0.014645 
>

RonL

**mr fantastic** · May 14th 2008, 05:59 AM

Originally Posted by CaptainBlack

That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.

So we can bust a gut attempting to prove that $\int_{-\infty}^{\infty} f(x)~dx \ne 1$ , but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $1$ so $f$ is not a density.

[snip]

Indeed.

And the given references don't say the product is a pdf either. They merely give an expression for the product. The expression is used to facilitate the proofs of other things.

**mr fantastic** · May 14th 2008, 06:18 AM

Originally Posted by CaptainBlack

That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian.
[snip]

Ha ha. I was going to take that route ....... Just to add some flesh to the idea:

$FT[\, f \cdot g\, ] = FT[\, f \, ] * FT[\, g \, ] =$ gaussian * gaussian = gaussian.

$FT[\, f \cdot g\, ] =$ gaussian therefore $f \cdot g =$ gaussian.

**vioravis** · May 26th 2008, 09:18 AM

Thanks a lot both. It was very useful.

**vioravis** · May 27th 2008, 08:51 PM

I have a couple of more questions on this product of two distributions:

1. In the derivation given by mr. fantastic, constant C has been ignored why defining mu and sigma? I am also not sure how can we define mu and sigma if the resultant expression is not a pdf?

2. Is this extensible for the multivariate case? I would appreciate if you can provide me some references in this regard.

Thanks a lot.

**mr fantastic** · May 27th 2008, 08:57 PM

Originally Posted by vioravis

I have a couple of more questions on this product of two distributions:

1. In the derivation given by mr. fantastic, constant C has been ignored why defining mu and sigma? I am also not sure how can we define mu and sigma if the resultant expression is not a pdf?

2. Is this extensible for the multivariate case? I would appreciate if you can provide me some references in this regard.

Thanks a lot.

They are just parameters in the Gaussians. Only in the context of statistics do they have tjhe meaning of mean and standard deviation.

C is ignored because it just becomes a multiplying factor: e^C = constant.

**aobosong** · March 1st 2009, 07:57 PM

Hi I have a problem similar to the poster's

f1(X) = {1/sigma1*sqrt(2*pi}* exp{-( y1-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-( y2-mu2^2)/2* sigma2^2}

notice the y1 and y2

and now y1 = x + n1
and y2 = x + n2
and n1 and n2 are normal distributions

I need to find the product of two functions and prove x follows a normal distribution...

Thank you.

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