# Thread: conditional probability

1. ## conditional probability

Suppose that independent random variables
X and Y have distributions N(4, 11) and

N
(6, 14), respectively.
(
a) Determine c such that Pr(X c) = 2Pr(X <c).
(
b) Find Pr{

4
(X Y )2 16}.
(
c) Suppose that Z is uniformly distributed over the interval [a, b]. Determine the

values of
a and b so that E(Y X) = E(Z) and Var(Y X) = Var(Z).

I can't do part a or part c. If you could go through either in stages it would be most appreciated. Thanks

2. a) $P(X>c) = 2P(X

Since $X \sim N(4,11)$, $Z = (X - 4)/\sqrt{11} \sim N(0,1)$
and then
$P(X
Now you have to look for a standarized normal table and look
for which the value of z such that P(Z < z) = 1/3.

b) Here you just have to find the distribution of W = X - Y. It can be done
using the Jacobian method, for example.

c) From letter b) you know what is the distribution of W, so
its expectation value and variance are also known. The formula to
compute the expectation and variance of a uniform variable are easily
obtained and then you have two equations with two unkown variables a and b.
Solve the system.

3. thanks i think that helped alot.

For part c) do you mean:

I use general formula's 0.5(a+b)=E(x)
& (1/12)(b-a)^2 = V(x)

to get two simultaneous equations, and then find values for a and c ?

Does it not matter that part b asked for X-Y, where as part c asks Y-X ?

4. Sorry, I didn't notice that in c) it is Y - X.

Actually, it is much simpler than what I have said.
You can use the fact that
E(X + Y) = E(X) + E(Y)
and
VAR(X + Y) = Var(X) + Var(Y)
since X and Y are independents, and use the formulas you described.

5. cheers for that

anychance you know how to answer part b and c of this:

Suppose that a random variable X has distribution function
F(x) :
0 - x< 2,
x/4-0.5 - 2 ≤ x ≤ 6,
1 - x> 6.
(i) Sketch the distribution function of X and find Pr(X >3).
(ii) Find the density function of X.
(iii) Find the mean and variance of X.

Part (i) is obvious
Part (ii) , would the answer be: f(x): 1/4 - (2 ≤ x ≤ 6) , 0 elsewhere
Part (iii) i don't really understand, even though i've got the formula's

Thanks again if you could help.

6. Yes, I think so.

For part lll remember that expectation and variance are just integrals. Since you know the density, just integrate xf(x) under the domain of X to obtain the expectation value (or mean). The variance can be obtained using the appropriate formula.

See you!