I have a v simple problem. If X(t) is standard brownian motion with mean = 0 and variance =1.Then, prove that c X(t/c^2) is a brownian motion as well. Let s < t.I can prove that mean of c * (X(t/c^2) - X (s/c^2) = 0 Also, variance of c * (X(t/c^2) - X (s/c^2) = t-s.But as simple as it sounds, I am not able to prove the moment generating function of c * X(t/c^2) to be guassian. This is where I get stuck. M.g.f cX(t/c^2) = M.g.f ( 1/c*X(t)), where c is scalar. But this does not show that variance of c.X(t/c^2) = t. I am some stupid mistake somewhere. Please comment.

i think it follows from definition X(t) - X(0) ~ N(0,t).. X(t/c^2) - X(0) ~ N(0,t/c^2), from here we can compute dP(cX(t/c^2) -cX(0) <=z)/dz (even this is not needed.. the specific corollary for normal variates can be used if eps~N(mu, sig^2) then k*eps~N(k*mu, (k*sig)^2) ))

QuoteOriginally posted by: mahalekrishnai think it follows from definition X(t) - X(0) ~ N(0,t).. X(t/c^2) - X(0) ~ N(0,t/c^2), from here we can compute dP(cX(t/c^2) -cX(0) <=z)/dz (even this is not needed.. the specific corollary for normal variates can be used if eps~N(mu, sig^2) then k*eps~N(k*mu, (k*sig)^2) ))Actually, my question is more basic. I am trying to prove that cX(t/c^2) is a guassian process give that X(t) is guassian. I understand it is a scalar multiplied with X(t/c^2) but I would like to prove from scratch. Can you help me?

QuoteOriginally posted by: iliketopologyQuoteOriginally posted by: mahalekrishnai think it follows from definition X(t) - X(0) ~ N(0,t).. X(t/c^2) - X(0) ~ N(0,t/c^2), from here we can compute dP(cX(t/c^2) -cX(0) <=z)/dz (even this is not needed.. the specific corollary for normal variates can be used if eps~N(mu, sig^2) then k*eps~N(k*mu, (k*sig)^2) ))Actually, my question is more basic. I am trying to prove that cX(t/c^2) is a guassian process give that X(t) is guassian. I understand it is a scalar multiplied with X(t/c^2) but I would like to prove from scratch. Can you help me?okay, lets hav a look at this onelet m(t) = cX(t/c^2)X(t) - X(s) ~ N(0, t-s)... this is just definition (mere representation)hence, X(t/c^2) - X(s/c^2) ~ N(0, (t-s)/c^2).. for the moment lets assume c>0P((m(t)-m(s))<z) = P( (X(t/c^2) - X(s/c^2)) < z/c) = integral {from -inf to z/c} (exp(-u^2/(2*(t-s)/c^2))/(2*pi*((t-s)/c^2) )^0.5) duNow to get the pdf of m(t) - m(s) .. just differentiate the integral w.r.t. to z.. for c<0 z/c will become the lower limit.. again the same thing, differentiate w.r.t. z

QuoteOriginally posted by: mahalekrishna okay, lets hav a look at this onelet m(t) = cX(t/c^2)X(t) - X(s) ~ N(0, t-s)... this is just definition (mere representation)hence, X(t/c^2) - X(s/c^2) ~ N(0, (t-s)/c^2).. for the moment lets assume c>0P((m(t)-m(s))<z) = P( (X(t/c^2) - X(s/c^2)) < z/c) = integral {from -inf to z/c} (exp(-u^2/(2*(t-s)/c^2))/(2*pi*((t-s)/c^2) )^0.5) duNow to get the pdf of m(t) - m(s) .. just differentiate the integral w.r.t. to z.. for c<0 z/c will become the lower limit.. again the same thing, differentiate w.r.t. zThanks. I think you proved it earlier also. I already obtained the mean and variance of c*X(t/c^2) as 0 and t respectively. Similarly, I proved the variance of c*(X(t/c^2) - X(s/c^2)) to be t-s. (0< s< t)As you mentioned, X(t) - X(s) ~ N(0, t-s).Hence, X(t/c^2) - X(s/c^2) ~ N(0, (t-s)/c^2). Therefore, c X(t/c^2) -c X(s/c^2) ~ N(0, (t-s)/). Essentially, you are saying that the pdf of say c X(t/c^2) that I used for generating m.g.f was wrong? Or else why would I get wrong answer by writing M.G.F of c*X(t/c^2) in terms of X(t)?I need to know also why this is wrong:m.g.f(c*x(t/c^2) = m.g.f(1/c*x(t)) = m.g.f (x (t/c)). This yields different answers altogethely? Now, although I proved that increments of cX(t/c^2) are normally distributed, I am stuck with my mfg being wrong.

QuoteOriginally posted by: iliketopologyQuoteOriginally posted by: mahalekrishna okay, lets hav a look at this onelet m(t) = cX(t/c^2)X(t) - X(s) ~ N(0, t-s)... this is just definition (mere representation)hence, X(t/c^2) - X(s/c^2) ~ N(0, (t-s)/c^2).. for the moment lets assume c>0P((m(t)-m(s))<z) = P( (X(t/c^2) - X(s/c^2)) < z/c) = integral {from -inf to z/c} (exp(-u^2/(2*(t-s)/c^2))/(2*pi*((t-s)/c^2) )^0.5) duNow to get the pdf of m(t) - m(s) .. just differentiate the integral w.r.t. to z.. for c<0 z/c will become the lower limit.. again the same thing, differentiate w.r.t. zThanks. I think you proved it earlier also. I already obtained the mean and variance of c*X(t/c^2) as 0 and t respectively. Similarly, I proved the variance of c*(X(t/c^2) - X(s/c^2)) to be t-s. (0< s< t)As you mentioned, X(t) - X(s) ~ N(0, t-s).Hence, X(t/c^2) - X(s/c^2) ~ N(0, (t-s)/c^2). Therefore, c X(t/c^2) -c X(s/c^2) ~ N(0, (t-s)/). Essentially, you are saying that the pdf of say c X(t/c^2) that I used for generating m.g.f was wrong? Or else why would I get wrong answer by writing M.G.F of c*X(t/c^2) in terms of X(t)?I need to know also why this is wrong:m.g.f(c*x(t/c^2) = m.g.f(1/c*x(t)) = m.g.f (x (t/c)). This yields different answers altogethely? Now, although I proved that increments of cX(t/c^2) are normally distributed, I am stuck with my mfg being wrong.why is the need to generate mgfs if you can prove the above using first principles.. and one more thing I didnt get your representation m.g.f(c*x(t/c^2).. i believe it is the moment generating function, if so it needs to be indexed using some other parameter (generally 't') not the rv itself

QuoteOriginally posted by: mahalekrishnaQuoteOriginally posted by: iliketopology why is the need to generate mgfs if you can prove the above using first principles.. and one more thing I didnt get your representation m.g.f(c*x(t/c^2).. i believe it is the moment generating function, if so it needs to be indexed using some other parameter (generally 't') not the rv itselfTrue, I don't need to but I know I am wrong with the mgf. Yes, that is moment generating function. I am trying to index properly but I might have to use subscript notation. I will try to write neater. Let MGF of X(u) = M(z), where u = t/c^2Hence, mean of X(u) is zero and variance is t/c^2 ( or u). Then MGF of cX(u) = M(cz) = exp(0*z *c + 0.5*variance(X)*z^2 * c^2) = exp( 0 + 0.5 * t/c^2*c^2*z^2) = exp( 0 + 0.5 * t z^2). Now, this is a guassian process with mean 0 and variance t. Hence, c*X(t/c^2) - c*X(0/c^2) is normally distributed. And so can the rest be proved. Here is where I made a mistake, Since X(t) has subcript "t", I combined that with the "t" used with the MGF. Not only that I made this mistake that MGF of X(t/c^2) can be written as MGF 1/c^2* X(t).This was totally absurd as you can see.