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This is the Exam of Probability which includes Density Function, Probability, Unbiased, Same Dice, Conditioning, Company, Construction Project etc. Key important points are: Compute the Probability, Interval, Density Function, Jointly Continuous, Marginal Probability, Conditional Probability, Probability, Correlation CoeCient, Number of Heads, Special Case
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MATH 4510: Review for Final Exam
We compute FX (x) = P (X ≤ x) = P (ln U ≤ x) = P (U ≤ ex) = e
x− 1 3 − 1 for 0^ ≤^ x^ ≤^ ln 3. So, the probability density function of X is
f (x) = d dx
F (x) =
ex 2 0 < x <^ ln 3 0 otherwise
So, P (X ≤ 1) =
0 f^ (x)dx^ =^
ex 2
e− 1
The random variables X and Y are jointly continuous with joint probability density function f (x, y) = cxy for 0 < x, y < 1 and f (x, y) = 0 otherwise. Find the constant c and compute the marginal probability density functions fX and fY. Find the conditional probability density function of X given that Y = y. Find P (XY ≤ 0 .5).
We have that 1 =
0
0 cxy^ dxdy^ =^
c 4 , so that^ c^ = 4. We compute:
fX (x) =
0
4 xydy = 2x for 0 < x < 1
fY (y) =
0
4 xydx = 2y for 0 < y < 1, and
fX|Y (x|y) =
f (x, y) fY (y)
4 xy 2 y
= 2x for 0 < x < 1.
Finally,
xy≤ 0. 5
4 xy dA =
0
0
4 xy dydx +
∫ (^21) x
0
4 xy dydx
0
2 x dx +
2 x 4 x^2
dx = x^2
ln x 2
ln 2 2
Aside: We see that f (x, y) = fX (x)fY (y), so that X and Y are independent, so we could have instead computed fX|Y by noting that fX|Y (x|y) = fX (x) when X and Y are independent.
We have that X and Y are both binomial with parameters n and p, so if we let σ^2 = np(1 − p), then var(X) = var(Y ) = σ^2. Also, X + Y = n, so
cov(X, Y ) = cov(X, n − X) = cov(X, n) − cov(X, X) = − var(X) = −σ^2.
Then, ρ(X, Y ) = cov(X, Y ) √ var(X) var(Y )
−σ^2 σ^2
Note that this is a special case of the fact that ρ(X, Y ) = ±1 if and only if Y = aX + b for some real numbers a and b with a 6 = 0.
Let Xi = 1 if the die lands on side i at least once and 0 otherwise. Let X be the num- ber of different sides that the die lands on, so that X = X 1 + · · · + X 6. We compute
E(Xi) = P (Xi = 1) = 1 − P (no roll lands on i) = 1 −
Then, E(X) =
i=1 E(Xi) = 6
6
Let Xi be the amount of time that Alex spends on the ith task and X = X 1 + · · · + X 10 be the amount of time that Alex spends on all of the tasks combined. Then, by the Central Limit Theorem,