Difficult Question Help (Probability/Statistics) Joint Probability Distributions

Hi Guys,

Hope everyone is well. I'm wondering if someone can help me with a difficult problem presented during lecture today.

Two friends plan to meet to go to a nightclub. Each of them arrives at a time uniformly distributed between midnight and 1am and independently of the other. Denote by X (respectively Y) the random variable representing the arrival time of the first person (respectively, the second). The joint probability distribution is given by

f_(x,y) (x,y) = 2 if 0 ≤ x ≤ y, and 0 otherwise.

a) Find the probability that the first person is waiting for his friend for more than 10 minutes.

b) Determine the marginal probability density functions of X and Y. Check that they are indeed probability density functions.

c) Calculate the means E[X] and E[Y].

d) Calculate the variances V(X) and V(Y).

e) Find the conditional density function of X given that Y = y, for 0 ≤ y ≤ 1. Check that it is indeed a probability density function.

f) Repeat part (e) for the conditional density of Y given that X = x.

Re: Difficult Question Help (Probability/Statistics) Joint Probability Distributions

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Originally Posted by

**Radiance** Hi Guys,

Hope everyone is well. I'm wondering if someone can help me with a difficult problem presented during lecture today.

Two friends plan to meet to go to a nightclub. Each of them arrives at a time uniformly distributed between midnight and 1am and independently of the other. Denote by X (respectively Y) the random variable representing the arrival time of the first person (respectively, the second). The joint probability distribution is given by

f_(x,y) (x,y) = 2 if 0 ≤ x ≤ y, and 0 otherwise.

a) Find the probability that the first person is waiting for his friend for more than 10 minutes.

b) Determine the marginal probability density functions of X and Y. Check that they are indeed probability density functions.

c) Calculate the means E[X] and E[Y].

d) Calculate the variances V(X) and V(Y).

e) Find the conditional density function of X given that Y = y, for 0 ≤ y ≤ 1. Check that it is indeed a probability density function.

f) Repeat part (e) for the conditional density of Y given that X = x.

Assuming $\displaystyle y \leq 1$ (initial problem did not state this, but it's safe to assume).

a) Since 10 min means X=1/6, we want to find $\displaystyle P(X>1/6)=1-P(x<1/6) = 1-\int_0^{1/6} \int_0^x f(x,y) dy dx$

b)$\displaystyle f_X(X) = \int_0^x f(x,y) dy$ and $\displaystyle f_Y(Y) = \int_0^y f(x,y) dx$, check that for both f's that the probability functions (the integrals of the f's) are equal to 1.

Once you have a) and b), calculating c)-f) should be straightforward enough ... you'll use the results of a) and b) (just use the definitions of expectancy, variance, conditional prob).