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suppose $$dA = \mu Adt + \sigma AdX.$$ is a geometric Brownian motion. One says that the Probability $P(A,t)$ of $A$ reashing the critical level $K(t)$ before maturity: $$\dfrac{\partial P}{\partial t} + \dfrac{1}{2}\sigma^2A^2\dfrac{\partial^2 P}{\partial A^2}+\mu \dfrac{\partial P}{\partial A} = 0$$ I know this is the Kolomogorov backward equation for transition density, but why this is true for probability function here? It seems we should replace CDF by PDF i.e replace $P$ by $\dfrac{\partial P}{\partial x}$?

Actually the background is the probability of default, Paul Wilmott on Quantitative Finance Volumne II Page641

suppose $$dA = \mu Adt + \sigma AdX.$$ is a geometric Brownian motion. One says that the Probability $P(A,t)$ of $A$ reashing the critical level $K(t)$ before maturity: $$\dfrac{\partial P}{\partial t} + \dfrac{1}{2}\sigma^2A^2\dfrac{\partial^2 P}{\partial A^2}+\mu \dfrac{\partial P}{\partial A} = 0$$ I know this is the Kolomogorov backward equation for transition density, but why this is true for probability function here? It seems we should replace CDF by PDF i.e replace $P$ by $\dfrac{\partial P}{\partial x}$?

Actually the background is the probability of default, Paul Wilmott on Quantitative Finance Volumne II Page641 39.2.2

suppose $$dA = \mu Adt + \sigma AdX.$$ is a geometric Brownian motion. One says that the Probability $P(A,t)$ of $A$ reashing the critical level $K(t)$ before maturity: $$\dfrac{\partial P}{\partial t} + \dfrac{1}{2}\sigma^2A^2\dfrac{\partial^2 P}{\partial A^2}+\mu \dfrac{\partial P}{\partial A} = 0$$ I know this is the Kolomogorov backward equation for transition density, but why this is true for probability function here?

Actually the background is the probability of default, Paul Wilmott on Quantitative Finance Volumne II Page641

suppose $$dA = \mu Adt + \sigma AdX.$$ is a geometric Brownian motion. One says that the Probability $P(A,t)$ of $A$ reashing the critical level $K(t)$ before maturity: $$\dfrac{\partial P}{\partial t} + \dfrac{1}{2}\sigma^2A^2\dfrac{\partial^2 P}{\partial A^2}+\mu \dfrac{\partial P}{\partial A} = 0$$ I know this is the Kolomogorov backward equation for transition density, but why this is true for probability function here? It seems we should replace CDF by PDF i.e replace $P$ by $\dfrac{\partial P}{\partial x}$?

Actually the background is the probability of default, Paul Wilmott on Quantitative Finance Volumne II Page641

suppose $$dA = \mu Adt + \sigma AdX.$$ is a geometric Brownian motion. One says that the Probability $P(A,t)$ of $A$ reashing the critical level $K(t)$ before maturity: $$\dfrac{\partial P}{\partial t} + \dfrac{1}{2}\sigma^2A^2\dfrac{\partial^2 P}{\partial A^2}+\mu \dfrac{\partial P}{\partial A} = 0$$ I know this is the Kolomogorov backward equation for transition density, but why this is true for probability function here?

Actually the background is the probability of default.

suppose $$dA = \mu Adt + \sigma AdX.$$ is a geometric Brownian motion. One says that the Probability $P(A,t)$ of $A$ reashing the critical level $K(t)$ before maturity: $$\dfrac{\partial P}{\partial t} + \dfrac{1}{2}\sigma^2A^2\dfrac{\partial^2 P}{\partial A^2}+\mu \dfrac{\partial P}{\partial A} = 0$$ I know this is the Kolomogorov backward equation for transition density, but why this is true for probability function here?

Actually the background is the probability of default, Paul Wilmott on Quantitative Finance Volumne II Page641