# Function of a random variable and conditioning

1. Oct 11, 2011

### 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

2. Oct 11, 2011

### Stephen Tashi

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 $\sigma$.

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?

Last edited: Oct 11, 2011
3. Oct 11, 2011

### Hejdun

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

4. Oct 11, 2011

### 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.)

5. Oct 11, 2011

### Hejdun

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.

6. Oct 12, 2011

### Stephen Tashi

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

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 $W$ ?