- Thread starter
- #1

The solution to this problem is -0.364, but I'm not sure how to approach this answer.

- Thread starter dlee
- Start date

- Thread starter
- #1

The solution to this problem is -0.364, but I'm not sure how to approach this answer.

- Admin
- #2

- Mar 5, 2012

- 8,887

I we assume a bivariate normal distribution, we "expect" the relation:

The solution to this problem is -0.364, but I'm not sure how to approach this answer.

$$y(x) = \text{sgn}(\rho) \frac {\sigma_Y}{\sigma_X} (x - \mu_X) + \mu_Y$$

With X at the 30th percentile, that means $z_X = \frac{x - \mu_X}{\sigma_X} = \text{invNorm}(0.30) = -0.524$.

In other words, the z-score for Y is

$$z_Y = \frac{y - \mu_Y}{\sigma_Y} = \text{sgn}(\rho) z_X = -0.524$$

I don't know how they got to -0.364.

That can't be right.I we assume a bivariate normal distribution, we "expect" the relation:

$$y(x) = \text{sgn}(\rho) \frac {\sigma_Y}{\sigma_X} (x - \mu_X) + \mu_Y$$

With X at the 30th percentile, that means $z_X = \frac{x - \mu_X}{\sigma_X} = \text{invNorm}(0.30) = -0.524$.

In other words, the z-score for Y is

$$z_Y = \frac{y - \mu_Y}{\sigma_Y} = \text{sgn}(\rho) z_X = -0.524$$

I don't know how they got to -0.364.

You can without loss of generality assume \(\displaystyle \mu_X = \mu_Y = 0\), so we have a model:

$$y=\alpha x$$

then $\displaystyle \sigma_Y=\alpha\; \sigma_X$, and $\rho=E(XY)/(\sigma_X \sigma_Y)=\alpha\; \sigma_X/\sigma_Y$

Hence: $$\alpha=\rho \frac{\sigma_Y}{\sigma_X}$$...

.

- Admin
- #4

- Mar 5, 2012

- 8,887

I didn't.That can't be right.

You can without loss of generality assume \(\displaystyle \mu_X = \mu_Y = 0\)

The problem asks for a z-score, meaning $\mu_X$, and $\mu_Y$ get eliminated (see my derivation).

Well... multiplying by 0.7so we have a model:

$$y=\alpha x$$

then $\displaystyle \sigma_Y=\alpha\; \sigma_X$, and $\rho=E(XY)/(\sigma_X \sigma_Y)=\alpha\; \sigma_X/\sigma_Y$

Hence: $$\alpha=\rho \frac{\sigma_Y}{\sigma_X}$$...

But that won't be right.

It will be if you use "nearest value" in inverse normal lookup in a table.... Well... multiplying by 0.7almostgives the requested result.

But that won't be right.

.

Last edited:

Well, since you failed to set up a model with the correct correlation it is not irrelevant to make an observation that simplifies setting the correlation without changing the answer.I didn't.

The problem asks for a z-score, meaning $\mu_X$, and $\mu_Y$ get eliminated (see my derivation).

.

Last edited:

- Admin
- #7

- Mar 5, 2012

- 8,887

The model is a positive sloped football that could be anywhere.Well, since you failed to set up a model with the correct correlation it is not irrelevant to make an observation that simplifies setting the correlation without changing the answer.

.

The problem puts the heart at x=4 with a variance of 4.

The y coordinate of the heart and the slope can still be freely chosen.

Then, with the given correlation, the "width" of the football becomes fixed.

Either way, when talking about the z-score of y, all these choices become moot, since they are standardized.

The relationship between $E(z_Y|z_X)$ and $z_X$ is simply $E(z_Y|z_X) = z_X$, whichever model you pick.

This is a "standardized" football that is aligned on the line y=x with a width such that the correlation is satisfied.

Last edited:

Since for Bivariate normal rv $X,\ Y$:The model is a positive sloped football that could be anywhere.

The problem puts the heart at x=4 with a variance of 4.

The y coordinate of the heart and the slope or can still be freely chosen.

Then, with the given correlation the "width" of the football becomes fixed.

Either way, when talking about the z-score of y, all these choices become moot, since they are standardized.

The relationship between $E(z_Y|z_X)$ and $z_X$ is simply $E(z_Y|z_X) = z_X$, whichever model you pick.

This is a "standardized" football that is aligned on the line y=x with a width such that the correlation is satisfied.

$$E(Y|X)=\rho\; \frac{\sigma_Y}{\sigma_X}Y$$

So as $z_X,\ z_Y$ have the same correlation coefficient as $X$ and $Y$ we have:

$$E(z_Y|z_X) = \rho\; z_X$$

See: http://athenasc.com/Bivariate-Normal.pdf.

... And simulation confirms this.

.

Last edited: