Function of a random variable and conditioning

[Hejdun]

Ok, since nobody answered my last problem, I simplify. :)

Let Z = γ1X1 + γ2X2, where the gammas are just constants
p(Z) = exp(Z)/(1 + exp(Z))
X1 and X2 are bivariate normal and put
Y = α + β1X1 + β2X2 + ε where ε ~ N(0,σ).

Now, we want to find f(p(Z)|X1,Y). In this case, is it legal to do the
operation f(p(Z)|X1,Y)=f(p(Z|X1,Y))?

That is can we write
f(exp(Z|X1,Y)/(1 + exp(Z|X1,Y)))?

Thanks for any help!
/H

 

[Stephen Tashi]

Let Z = γ1X1 + γ2X2, where the gammas are just constants
p(Z) = exp(Z)/(1 + exp(Z))
X1 and X2 are bivariate normal and put
Y = α + β1X1 + β2X2 + ε where ε ~ N(0,σ).

Let $X_1$ and $X_2$ each be bivariate normal random variables
Let $Z = \gamma_1 X_1 + \gamma_2 X_2$ where $\gamma_1$ and $\gamma_2$ are constants.
Let $Y = \alpha + \beta_1 X_1 + \beta_2 X-2 + \epsilon$ where $\alpha,\beta_1,\beta_2$ are each constants and $\epsilon$ is a normal random variable with mean 0 and standard deviation $\sigma$.

Now, we want to find f(p(Z)|X1,Y).

Is that notation supposed mean you want the probability density function for $Z$ given $X_1$ and $Y$ ?

In this case, is it legal to do the
operation f(p(Z)|X1,Y)=f(p(Z|X1,Y))?
That is can we write
f(exp(Z|X1,Y)/(1 + exp(Z|X1,Y)))?

I don't know what that notation means. The conditional density function of $Z$ is some function of the variables $Z, X_1,Y$ but what does the notation "exp(Z|X1,Y)" mean?

 

[Hejdun]

Is that notation supposed mean you want the probability density function for $Z$ given $X_1$ and $Y$ ?

I don't know what that notation means. The conditional density function of $Z$ is some function of the variables $Z, X_1,Y$ but what does the notation "exp(Z|X1,Y)" mean?

Yes.


For instance, if we want to know the distribution of p(Z) = exp(Z)/(1 + exp(Z)) and we know the distribution of Z, then we make a simple transformation, put the inverse in the pdf of Z and multiply with the derivative as usual.

However, the problem is finding the disitrbution of p(Z)|Y. My idea was then to put the inverse in the pdf of Z|Y and then multiply with the inverse. I am not sure if my approach is correct, but if you have another suggestion of how to proceed I would be grateful.

/H

 

[Stephen Tashi]

Let's try again:

Let $X_1$ and $X_2$ be random variables that have a joint bivariate normal distribution (rather than each of them being bivariate normal).
Let $Z = \gamma_1 X_1 + \gamma_2 X_2$ where $\gamma_1$ and $\gamma_2$ are constants.
Let $W = \exp(Z)/(1 + \exp(Z))$
Let $Y = \alpha + \beta_1 X_1 + \beta_2 X_2 + \epsilon$ where $\alpha,\beta_1,\beta_2$ are each constants and $\epsilon$ is a normal random variable with mean 0 and standard deviation $\sigma$.

Do you want the conditional distribution of $W$ given $X_1$ and $Y$ ? (Your other thread mentioned a joint distribution instead of conditional distribution and also it mentioned that the final goal was to find an expected value.)

 

[Hejdun]

Let's try again:

Let $X_1$ and $X_2$ be random variables that have a joint bivariate normal distribution (rather than each of them being bivariate normal).
Let $Z = \gamma_1 X_1 + \gamma_2 X_2$ where $\gamma_1$ and $\gamma_2$ are constants.
Let $W = \exp(Z)/(1 + \exp(Z))$
Let $Y = \alpha + \beta_1 X_1 + \beta_2 X_2 + \epsilon$ where $\alpha,\beta_1,\beta_2$ are each constants and $\epsilon$ is a normal random variable with mean 0 and standard deviation $\sigma$.

Do you want the conditional distribution of $W$ given $X_1$ and $Y$ ? (Your other thread mentioned a joint distribution instead of conditional distribution and also it mentioned that the final goal was to find an expected value.)

The final goal is to find the conditional distribution of X1 and Y given W. Of course, there are different ways of getting there depending on how you calculate the joint X1, Y, W.

My question in this thread may solve a part of the problem and also the disitrbution of W given X1 and Y.

 

[Stephen Tashi]

Now that the problem is established, help me understand the question about technique.

For instance, if we want to know the distribution of p(Z) = exp(Z)/(1 + exp(Z)) and we know the distribution of Z, then we make a simple transformation, put the inverse in the pdf of Z and multiply with the derivative as usual.

Who's inverse and who's derivative are you talking about? Let's say $Z$ has cumulative distribution $F_Z(x)$ with inverse function ${F_Z}^{-1}(x)$. Using that notation, what is your claim about the probability density (or cumulative distribution) of $W$ ?