# Guassian pdf is sum of two other RV Integral evaluation -

1. Jul 11, 2011

Could someone please advise me how to put the integral from + to - infinity notation used $exp{-1/2(y-x)^2(A^-1) + x^2(B)^-1}dx Into the form k1$ exp[-k2(x-k3)^2]dx using the fact that $exp(-1/2x^2)dx = 2pi^1/2 I have encountered this problem when trying to derive the conditional density function of p(x|y) where y = x + n, where y is a gaussian random variable and is the sum of two other random variables. Help really appreciated with this - yhanks - - Ems x 2. Jul 11, 2011 ### pmsrw3 Just substitute x+n for y, expand, collect terms, and complete the square. 3. Jul 12, 2011 ### EmmaSaunders1 Hi Thanks for the help - when make the substitution I obtain$exp{-1/2(n)^2(A^-1) + x^2(B^-1)} - I am having trouble putting into the form k1 $exp[-k2(x-k3)^2]. As it is now I would have an extra term of -2nx which doesnt fit the above - your help is appreciated. Thanks Emma x 4. Jul 12, 2011 ### pmsrw3 That's not right. There should be an -1/2 x^2(A^-1) term, and a -2nx term, too. You have inside the exp an expression of the form $a x^2 + b x + c$, where a, b, and c are combinations of n, A, and B. You now need to complete the square -- that means find a number d such that $a x^2 + b x + d = (x\sqrt{a}+\sqrt{d})^2$. You'll find that $d = \frac{b^2}{4a}$ works. Now you can write: $$\begin{eqnarray*} a x^2 + b x + c & = & a x^2 + b x + \frac{b^2}{4a} - \frac{b^2}{4a} + c \\ & = & \left(x\sqrt{a} + \frac{b}{2\sqrt{a}} \right)^2 - \frac{b^2}{4a} + c \\ & = & a \left(x + \frac{b}{2a} \right)^2 + \left( c -\frac{b^2}{4a} \right)\\ \end{eqnarray*}$$ (I may have made a mistake or two in there, but that's the strategy.) You can read k2 and k3 off the first term. k1 is the exponential of the last thing. Actually, your 1/n^2 term will contribute to k1, too. This should look familiar -- it's the way the quadratic equation $a x^2 + b x + c = 0$ is solved. 5. Jul 12, 2011 ### EmmaSaunders1 Thanks for your quick response. When making the substitution however, considering the first terms of A^-1, we get = 1/2{x^2+n^2+2xn+x^2-2x^2-2xn]A^-1 which all cancel to 1/2[n^2]A^-1. Are you leaving this expression in its full form or not? Thanks again - Emmax 6. Jul 12, 2011 ### pmsrw3 You're right -- my error. That makes the problem considerably simpler -- you don't need to complete the square. But what did you mean then, by saying, "I would have an extra term of -2nx"? 7. Jul 12, 2011 ### EmmaSaunders1 if after expansion and collection of terms - I get exp{-1/2(n)^2(A^-1) + x^2(B^-1). I cant see how to put this into the form exp[-k2(x-k3)^2]. Expanding the latter out results in terms of x^2 +k3^2 - 2k3x. This last term throws me off - if you can see this simpler I d appreciate your oppinion - thanks x 8. Jul 12, 2011 ### pmsrw3 Well, in your case k3 = 0. 9. Jul 12, 2011 ### EmmaSaunders1 But then you wont have an n^2 term if expanding 10. Jul 12, 2011 ### pmsrw3 Maybe this is what you're missing: $$e^{\frac{x^2}{B}-\frac{n^2}{2 A}}=e^{\frac{x^2}{B}}e^{-\frac{n^2}{2 A}}=k_1e^{\frac{x^2}{B}}$$ Does that help? 11. Jul 12, 2011 ### EmmaSaunders1 possibly - could you please clarify the notation - - frac? 12. Jul 12, 2011 ### pmsrw3 Ah! I see you're using a browser that doesn't display LaTex equations properly. Well, that will have made all my previous posts infinitely more difficult to read. \frac{numerator}{denominator} means numerator/denominator. 13. Jul 12, 2011 ### EmmaSaunders1 Thanks - I will have a look tonight and try to update my browser - could you possibly clarify - if I have an equation of the form Ax^2 + bx + c , how do I turn it into the form k2(x-k3)^2 14. Jul 12, 2011 ### pmsrw3 15. Jul 12, 2011 ### EmmaSaunders1 Have you found the terms k1, k2, k3? I am thinking it may be easier without the substitution of y = x+n. Thanks for your help x 16. Jul 12, 2011 ### EmmaSaunders1 Performing expansion and collection of terms I get - exp[-1/2{x^2(A^-1+B^-1)+n^2(A^-1)} - - using the n term substitution. Using no substitution I get exp[-1/2{x^2(A^-1+B^-1) - 2yxA^-1 + y^2A^-1] = which looks more like solving the square problem - but I still cant work out how to find the terms k1, k2 or k3 from this? 17. Jul 12, 2011 ### pmsrw3 18. Jul 12, 2011 ### pmsrw3 Uhh... Now you're REALLY confusing me. You said earlier, and convinced me you were right, that you got exp{x^2(B^-1)+n^2((2A)^-1))}. Why did it change? 19. Jul 12, 2011 ### EmmaSaunders1 When I dont use the original y=x+n substitution I get [-1/2{x^2(A^-1+B^-1) - 2yxA^-1 + y^2A^-1] - then substituting y for x+n - I find exp[-1/2{x^2(A^-1+B^-1)+n^2(A^-1)} . I appreciate your help. The problems solution ultimately needs to be in terms of x and y. It is involved around evaluating the integral to find the posterior distribution for two gaussian rv's one of which is the sum (i.e y = x+ n) of another two rv's. When working through the problem it seems that k1 is of the form AB/(A+B) 20. Jul 12, 2011 ### pmsrw3 I don't see how this can possible be true if you're starting from exp{-1/2(y-x)^2(A^-1) + x^2(B)^-1} as you said in the original post. Anyway, "[-1/2{x^2(A^-1+B^-1) - 2yxA^-1 + y^2A^-1]" is wrong: the "{" is unbalanced. What is being multiplied by -1/2 here? Well, if you say that's what you find, I can't say it's untrue. But if that's what you find, you've made a mistake. The -2yx/A term contributes a -2x^2/A that cancels out the x^2/A terms. Emma, you said in your original post that you wanted a solution in the form "k1$ exp[-k2(x-k3)^2]dx". There is no y in there. Why do you now say "The problems solution ultimately needs to be in terms of x and y?

What the Hell is an rv? I assume you don't mean "Recreational Vehicle".

Well, if so, then you must have given some incorrect information, because that is not the answer to the question you posed.

I suspect that you're supposed to normalize this distribution, and that that's where an AB/(A+B) factor might come in. However, there was no way to figure this out from what you've said so far. Honestly, I'm getting pretty frustrated. You keep giving inaccurate, incomplete, and inconsistent information, in a notation that is almost unreadable. Couldn't you be a little more careful?