Q6c

Now, 
f(y,t|x,s)=(∂/(∂y))(P((Subscript[Z, t])<y|(Subscript[Z, s])=x))
Thus, 
P((Subscript[Z, t])<y|(Subscript[Z, s])=x)
= P(exp((E^(-t))+(E^(-t))(∫_0^t)(E^u)\[DifferentialD](Subscript[B, u]))<y|exp((E^(-s))+(E^(-s))(∫_0^s)(E^u)\[DifferentialD](Subscript[B, u]))=x)
= P(exp((E^(-t)))exp((E^(-t))(∫_0^t)(E^u)\[DifferentialD](Subscript[B, u]))<y|exp((E^(-s)))exp((E^(-s))(∫_0^s)(E^u)\[DifferentialD](Subscript[B, u]))=x)
= P(exp((E^(-s))(E^(-(t-s))))exp((E^(-s))(E^(-(t-s)))(∫_0^s)(E^u)\[DifferentialD](Subscript[B, u]))exp((E^(-s))(E^(-(t-s)))(∫_s^t)(E^u)\[DifferentialD](Subscript[B, u]))<y|exp((E^(-s)))exp((E^(-s))(∫_0^s)(E^u)\[DifferentialD](Subscript[B, u]))=x)
= P((x^(E^(-(t-s))))exp((E^(-t))(∫_s^t)(E^u)\[DifferentialD](Subscript[B, u]))<y|exp((E^(-s)))exp((E^(-s))(∫_0^s)(E^u)\[DifferentialD](Subscript[B, u]))=x)
= P((x^(E^(-(t-s))))exp((E^(-t))(∫_s^t)(E^u)\[DifferentialD](Subscript[B, u]))<y) (by independence)
= P((E^(-(t-s)))ln(x)+(E^(-t))(∫_s^t)(E^u)\[DifferentialD](Subscript[B, u])<ln(y))
= P((E^(-t))(∫_s^t)(E^u)\[DifferentialD](Subscript[B, u]) < ln(y) - (E^(-(t-s)))ln(x))

Now,
(∫_s^t)((((E^u)))^2)\[DifferentialD]u=(∫_s^t)(E^(2u))\[DifferentialD]u
=(((E^(2t))-(E^(2s)))/2)<∞
So, (∫_s^t)(E^u)\[DifferentialD](Subscript[B, u])~N(0,(((E^(2t))-(E^(2s)))/2))
(E^(-t))(∫_s^t)(E^u)\[DifferentialD](Subscript[B, u])~N(0,(E^(-2t))(((E^(2t))-(E^(2s)))/2))
(E^(-t))(∫_s^t)(E^u)\[DifferentialD](Subscript[B, u])~N(0,((1-(E^(-2(t-s))))/2))
Thus,μ=0, (σ^2)=((1-(E^(-2(t-s))))/2)
So,
F(y,t|x,s)=(∫_(-∞)^(ln(y)-(E^(-(t-s)))ln(x)))(1/(Sqrt[2π(((1-(E^(-2(t-s))))/2))]))(E^(-((u^2)/(2(((1-(E^(-2(t-s))))/2))))))\[DifferentialD]u
=(∫_(-∞)^(ln(y)-(E^(-(t-s)))ln(x)))(1/(Sqrt[π(1-(E^(-2(t-s))))]))exp(-((u^2)/(1-(E^(-2(t-s))))))\[DifferentialD]u
Therefore,
f(y,t|x,s)=(1/(Sqrt[π(1-(E^(-2(t-s))))]))exp(-((((ln(y)-(E^(-(t-s)))ln(x)))^2)/(1-(E^(-2(t-s))))))-0
=(1/(Sqrt[π(1-(E^(-2(t-s))))]))exp(-((((ln(y)-(E^(-(t-s)))ln(x)))^2)/(1-(E^(-2(t-s))))))
Thus, shown.