With ITO FORMULA and u(t; x) = t we have : d(tBt) = Bt dt

With ITO FORMULA and u(t; x) = w we have : d(tBt) = Bt dt + t dBt.

With ITO FORMULA and u(t; x) = x we have : d(tBt) =  t dBt.

With ITO FORMULA and u(t; x) = tx we have : d(tBt) = Bt dt - t dBt.

With ITO FORMULA and u(t; x) = tx we have : d(tBt) = Bt dt + t dBt.

Xt follow a normal law with variance equal to : (1/3)t^3

Xt follow a centered normal law with variance equal to : (1/3)t^3

Xt follow a centered normal law with variance equal to : t^3

Xt follow a normal law with variance equal to : (1/3)t

Xt follow a centered normal law with variance equal to : 1

E(Xt)=0 and VAR(Xt)=(1/3) t^3

E(Xt)=0 and VAR(Xt)=1.

E(Xt)=0 and VAR(Xt)=t

E(Xt)=1 and VAR(Xt)=t.

E(Xt)=0 and VAR(Xt)=t^2

With ITO FORMULA, we find:

We take : u(t,x)=e^(-t) * x

dYt=- d Bt

With ITO FORMULA, we find:

We take : u(t,x)=e^(-t) * x

dYt=-Yt dt + d Bt

With ITO FORMULA, we find:

We take : u(t,x)=e^(-t)

dYt=-Yt dt

With ITO FORMULA, we find:

We take : u(t,x)=e^(t) * x

dYt=Yt dt + d Bt

With characteristic function, we find a normal law with variance equal to (1/2).

With characteristic function, we find a centered normal law with variance equal to (1/2).

With characteristic function, we find a centered normal law with variance equal to 1.

With characteristic function, we find a binomial law with parameter p equal to (1/2).

Xt and Yt follow centered normal law with variance 1.

Xt and Yt follow geometric law with parameter1.

Xt and Yt follow centered normal law with variance of question 1.

Xt and Yt follow binomial law with parameter 1.

Xt and Yt follow centered poisson law with parameter 1.

E(Xt)=1, Var(Xt)=(1/2), E(Yt)=1 and Var(Yt)=(1/2)

E(Xt)=0, Var(Xt)=(1/2) [e(2t) -1]., E(Yt) and Var(Yt)=0

(1/2) [1-e(-2t)]

E(Xt)=1, Var(Xt)=1, E(Yt)=0 and Var(Yt)=1

E(Xt)=0, Var(Xt)=0, E(Yt)=0 and Var(Yt)=0

E(Xt)=0, Var(Xt)=(1/2) [1-e(-2t)]

, E(Yt)=0 and Var(Yt)=(1/2) [e(2t) -1]