Conditional probability density function

[drullanorull]

Please help me with this. Any suggestions are greatly appreciated.
Imagine that I have a bank account. X is the amount of cash on the account at time t+1. Y is the amount of cash at time t. The amount of cash depends on the deposits made and on the amount of cash during the previous period. The deposits are made based on a random variable, Z, (stock returns) which has a probability density that is log-normal distributed. My question is what the probability density function for the amount of cash at time t+1 looks like.
$$f_X(x)=\int f_{X,Y}(x,y)dy = \int f_{X|Y}(x|y)f_Y(y)dy$$
My problem is how to relate $$f_{X|Y}(x|y)$$ with the deposits.
Is $$f_{X|Y}(x|y)=y+f_Z(z)$$
So that
$$f_X(x)=\int(y+f_Z(z))f_Y(y)dy$$

Is this true?

 

[Adeimantus]

Please help me with this. Any suggestions are greatly appreciated.
Imagine that I have a bank account. X is the amount of cash on the account at time t+1. Y is the amount of cash at time t. The amount of cash depends on the deposits made and on the amount of cash during the previous period. The deposits are made based on a random variable, Z, (stock returns) which has a probability density that is log-normal distributed. My question is what the probability density function for the amount of cash at time t+1 looks like.
$$f_X(x)=\int f_{X,Y}(x,y)dy = \int f_{X|Y}(x|y)f_Y(y)dy$$
My problem is how to relate $$f_{X|Y}(x|y)$$ with the deposits.
Is $$f_{X|Y}(x|y)=y+f_Z(z)$$
So that
$$f_X(x)=\int(y+f_Z(z))f_Y(y)dy$$

Is this true?

X = Y+Z. So given that Y=y, you want the density that X = y + Z = x, or Z = x-y. So
$$f_{X|Y}(x|y)=f_Z(x-y)$$