# conditional probability

• April 19th 2008, 12:11 PM
winganger
conditional probability
Suppose that the heights of a parent and child, P and C in cm are jointly Gaussian
with
mP = 170cm, standard deviation of P = 8cm
mC = 170cm, standard deviation of C = 8cm
correlation cofficient between P and C = 0.2
Find P[C>mC|P= mP], the probability that the child is above average height if the parent is average height.
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• April 19th 2008, 09:55 PM
mr fantastic
Quote:

Originally Posted by winganger
Suppose that the heights of a parent and child, P and C in cm are jointly Gaussian
with
mP = 170cm, standard deviation of P = 8cm
mC = 170cm, standard deviation of C = 8cm
correlation cofficient between P and C = 0.2
Find P[C>mC|P= mP], the probability that the child is above average height if the parent is average height.
(Clapping)(Clapping)

I have no time now but will take a look later if no-one else replies. In the meantime, did you intend mP and mC to both equal 170 cm? Similarly with the equal standard deviations?
• April 20th 2008, 02:21 AM
mr fantastic
Quote:

Originally Posted by winganger
Suppose that the heights of a parent and child, P and C in cm are jointly Gaussian
with
mP = 170cm, standard deviation of P = 8cm
mC = 170cm, standard deviation of C = 8cm
correlation cofficient between P and C = 0.2
Find P[C>mC|P= mP], the probability that the child is above average height if the parent is average height.
(Clapping)(Clapping)

There's the easy way and and there's the hard way. Being lazy by nature, I'll outline the key results for the easy way:

If X and Y are jointly gaussian (bivariate normal distribution), then the conditional pdf for one of the variables, given a known value for the other variable, is normally distributed:

$f(x | y = y_0) = \text{Normal} \left(\mu_{x|y = y_0}, \, \sigma^2_{x|y=y_0}\right)$

where:

$\mu_{x|y = y_0} = \mu_x + \rho \sigma_x \frac{(y_0 - \mu_y)}{\sigma_y}$

$\sigma_{x|y=y_0} = \sigma_x \sqrt{1 - \rho^2}$

and $\rho$ is the correlation coefficient.

I'll leave it to you to substitute the necessary notation and values. It should then be routine to calculate $\Pr(C > \mu_C | P = \mu_P)$.

• April 20th 2008, 03:00 AM
winganger
Quote:

Originally Posted by mr fantastic
There's the easy way and and there's the hard way. Being lazy by nature, I'll outline the key results for the easy way:

If X and Y are jointly gaussian (bivariate normal distribution), then the conditional pdf for one of the variables, given a known value for the other variable, is normally distributed:

$f(x | y = y_0) = \text{Normal} \left(\mu_{x|y = y_0}, \, \sigma^2_{x|y=y_0}\right)$

where:

$\mu_{x|y = y_0} = \mu_x + \rho \sigma_x \frac{(y_0 - \mu_y)}{\sigma_y}$

$\sigma_{x|y=y_0} = \sigma_x \sqrt{1 - \rho^2}$

and $\rho$ is the correlation coefficient.

I'll leave it to you to substitute the necessary notation and values. It should then be routine to calculate $\Pr(C > \mu_C | P = \mu_P)$.

Thank you. After substituted the values and I get the answer
N(0, 0.96) , required probability = 0.5
Could I ask one more question, i.e. how to find the probability Pr(C > mC | P > x)?
• April 20th 2008, 03:38 AM
mr fantastic
Quote:

Originally Posted by winganger
Thank you. After substituted the values and I get the answer
N(0, 0.96) , required probability = 0.5
Could I ask one more question, i.e. how to find the probability Pr(C > mC | P > x)?

$\Pr(Y > a | X > b) = \frac{\int_{a}^{\infty} \int_{b}^{\infty} f(x, y) \, dx \, dy}{\int_{b}^{\infty} g(x) \, dx}$

where f(x, y) is the joint pdf of X and Y, and g(x) is the marginal pdf of X.