Multivariate normal distribution , conditional probability problem

[wuid]

Homework Statement

Given X,Y RV both have normal distribution with:
[itex]μ_{x}=6[/itex],[itex]μ_{y}=4[/itex],[itex]σ_{x}=1[/itex],[itex]σ_{y}=5[/itex],ρ=0.1

a. are X,Y independent?
b. find P(X≤5)
c. find P(Y≤5|X=5)

2. The attempt at a solutiona. no -> ρ=0.1 -> cov(X,Y)≠0

b. define Z=[itex]\frac{X-6}{1}[/itex] ; Z~N(0,1)
so P(X≤5) = P(Z≤-1) = [itex]\Phi(-1)[/itex]

c. Here i have my difficulty were i able to calculate and i know :
μ=[6 4] , Ʃ=[1 0.5,0.5 25] cov matrix , [itex]Ʃ^{-1}[/itex]= [[itex]\frac{100}{99} \frac{-2}{99}, \frac{-2}{99} \frac{4}{99}[/itex]] inverse matrix.just don't how to deal with the conditional question...

 

[mighty2000]

c. To find P(Y≤5|X=5), we can use the formula for conditional probability:

P(Y≤5|X=5) = P(Y≤5 and X=5) / P(X=5)

To find the numerator, we can use the joint probability formula:

P(Y≤5 and X=5) = P(X=5) * P(Y≤5|X=5)

Since X and Y are normally distributed with means μx=6 and μy=4, respectively, and standard deviations σx=1 and σy=5, we can use the normal distribution formula to calculate the probabilities:

P(X=5) = \frac{1}{\sqrt{2\pi}} * e^{-\frac{1}{2}(\frac{5-6}{1})^2} = \frac{1}{\sqrt{2\pi}} * e^{-\frac{1}{2}(-1)^2} = \frac{1}{\sqrt{2\pi}} * e^{-\frac{1}{2}} = 0.241971

P(Y≤5|X=5) = \frac{1}{\sqrt{2\pi}} * e^{-\frac{1}{2}(\frac{5-4}{5})^2} = \frac{1}{\sqrt{2\pi}} * e^{-\frac{1}{2}(\frac{1}{5})^2} = \frac{1}{\sqrt{2\pi}} * e^{-\frac{1}{50}} = 0.398942

Therefore, the numerator is:

P(Y≤5 and X=5) = 0.241971 * 0.398942 = 0.096557

And the final answer is:

P(Y≤5|X=5) = \frac{0.096557}{0.241971} = 0.398942

So the probability of Y being less than or equal to 5, given that X is equal to 5, is approximately 0.398942 or 39.89%.