15 Problems Solved
Welcome to Tutorial 5! Here, we move beyond single variables to analyze relationships between two random variables. Pay close attention to double integrals for joint PDFs and summation for joint PMFs. Covariance and Correlation are key metrics you must master.
X and Y are proportions of time drive-in and walk-in facilities are in use. Joint density: \[ f(x,y) = \begin{cases} \frac{2}{5}(2x+3y) & 0 \le x,y \le 1 \\ 0 & \text{otherwise} \end{cases} \] Find \( P((X,Y) \in A) \) where \( A = \{(x,y) \mid 0 \le x \le 1/2, 1/4 \le y \le 1/2\} \).
Set up a double integral with limits \( x \in [0, 1/2] \) and \( y \in [1/4, 1/2] \).
Integrating over the region:
\[ \int_{1/4}^{1/2} \int_0^{1/2} \frac{2}{5}(2x+3y) \, dx \, dy \]
Inner integral w.r.t. \( x \):
\[ \frac{2}{5} [x^2 + 3xy]_0^{1/2} = \frac{2}{5}(1/4 + 3y/2) = \frac{1+6y}{10} \]
Outer integral w.r.t. \( y \):
\[ \int_{1/4}^{1/2} \frac{1+6y}{10} \, dy = \frac{1}{10} [y + 3y^2]_{1/4}^{1/2} \]
\[ = \frac{1}{10} [(1/2 + 3/4) - (1/4 + 3/16)] = \frac{1}{10} [5/4 - 7/16] \]
Find the marginal PDFs for random variables X and Y for the density function in Problem 1.
Integrate out the "other" variable over its entire range (0 to 1).
Marginal of X:
\[ g(x) = \int_0^1 f(x,y) \, dy = \int_0^1 \frac{2}{5}(2x+3y) \, dy = \frac{2}{5}[2xy + \frac{3y^2}{2}]_0^1 = \frac{4x+3}{5} \]
Marginal of Y:
\[ h(y) = \int_0^1 f(x,y) \, dx = \int_0^1 \frac{2}{5}(2x+3y) \, dx = \frac{2}{5}[x^2 + 3xy]_0^1 = \frac{2(1+3y)}{5} \]
3 Blue, 2 Red, 3 Green pens. Select 2 at random. X = number of Blue, Y = number of Red. Find (a) Joint PMF, (b) \( P(X+Y \le 1) \).
Use combination formula \(\binom{n}{r}\). Total ways = \(\binom{8}{2} = 28\).
(a) Joint PMF Table:
| (x,y) | Ways | P(X=x, Y=y) |
|---|---|---|
| (0,0) | \(\binom{3}{0}\binom{2}{0}\binom{3}{2}=3\) | 3/28 |
| (1,0) | \(\binom{3}{1}\binom{2}{0}\binom{3}{1}=9\) | 9/28 |
| (2,0) | \(\binom{3}{2}\binom{2}{0}\binom{3}{0}=3\) | 3/28 |
| (0,1) | \(\binom{3}{0}\binom{2}{1}\binom{3}{1}=6\) | 6/28 |
| (1,1) | \(\binom{3}{1}\binom{2}{1}\binom{3}{0}=6\) | 6/28 |
| (0,2) | \(\binom{3}{0}\binom{2}{2}\binom{3}{0}=1\) | 1/28 |
(b) \( P(X+Y \le 1) \):
\[ P = f(0,0) + f(1,0) + f(0,1) = \frac{3+9+6}{28} = \frac{18}{28} = \frac{9}{14} \]
In Problem 3, find the conditional distribution of X given Y=1. Calculate \( P(X=0 \mid Y=1) \).
Divide the joint probabilities by the marginal total \( P(Y=1) \).
\[ P(Y=1) = f(0,1) + f(1,1) = 6/28 + 6/28 = 12/28 = 3/7 \]
Conditional distribution of X for \( Y=1 \):
Determine \( E[XY] \) for the pens selection problem in Problem 3.
Only the term where both \( x,y > 0 \) contributes to the sum.
\[ E[XY] = \sum_x \sum_y xy \cdot f(x,y) \]
The only non-zero term is \( (1 \times 1) \times f(1,1) \):
\[ E[XY] = 1 \times 1 \times \frac{6}{28} = \frac{3}{14} \]
Calculate the covariance \( Cov(X,Y) \) for the pens selection problem.
Use formula: \( Cov(X,Y) = E[XY] - E[X]E[Y] \).
1. Find \( E[X] \): \( 1(\frac{9+6}{28}) + 2(\frac{3}{28}) = \frac{15+6}{28} = \frac{3}{4} \)
2. Find \( E[Y] \): \( 1(\frac{12}{28}) + 2(\frac{1}{28}) = \frac{14}{28} = \frac{1}{2} \)
3. \( Cov(X,Y) = \frac{3}{14} - (\frac{3}{4} \times \frac{1}{2}) = \frac{3}{14} - \frac{3}{8} \)
\[ Cov(X,Y) = \frac{12 - 21}{56} = -9/56 \]
Determine the correlation coefficient \(\rho_{XY}\) for the pens selection problem.
Divide covariance by the product of standard deviations.
1. \( Var(X) = E[X^2] - (E[X])^2 = 27/28 - 9/16 = 45/112 \)
2. \( Var(Y) = E[Y^2] - (E[Y])^2 = 4/7 - 1/4 = 9/28 \)
3. \( \rho = \frac{-9/56}{\sqrt{45/112 \times 9/28}} = -1/\sqrt{5} \)
Joint density: \( f(x,y) = 10xy^2, 0 < x < y < 1 \). (a) Find marginals and \( f(y|x) \). (b) \( P(Y> 1/2 \mid X = 0.25) \).
Watch the limits! \( y \) goes from \( x \) to 1, and \( x \) goes from 0 to \( y \).
(a) Densities:
\( g(x) = \int_x^1 10xy^2 \, dy = \frac{10}{3}x(1-x^3) \)
\( h(y) = \int_0^y 10xy^2 \, dx = 5y^4 \)
\( f(y|x) = \frac{3y^2}{1-x^3} \)
(b) Probability:
\( P(Y > 1/2 \mid X=0.25) = \int_{1/2}^1 \frac{3y^2}{1-(1/4)^3} \, dy = 8/9 \)
\( f(x,y) = \frac{x(1+3y^2)}{4}, 0 < x < 2, 0 < y < 1 \). Find \( f(x|y) \) and evaluate \( P(1/4 < X < 1/2 \mid Y=1/3) \).
Notice that the PDF can be factored into \( (x/2) \cdot ((1+3y^2)/2) \).
The variables are independent because \( f(x,y) = g(x)h(y) \).
\( f(x|y) = g(x) = x/2 \)
\[ P(1/4 < X < 1/2)=\int_{1/4}^{1/2} x/2 \, dx=[x^2/4]_{1/4}^{1/2}=3/64 \]
\( f(x,y) = 6x, 0 < x < 1, 0 < y < 1-x \). Prove/Disprove independence.
Look at the range. Does the range of Y depend on X?
Verification:
1. The range is triangular (\( y < 1-x \)).
2. For independence, the domain MUST be a rectangle.
3. Calculated Marginals: \( g(x) = 6x(1-x) \) and \( h(y) = 3(1-y)^2 \).
4. \( g(x)h(y) \neq f(x,y) \).
\( f(x,y) = 8xy, 0 < y < x < 1 \). Find the covariance of X and Y.
Use \( Cov(X,Y) = E[XY] - E[X]E[Y] \). Integrate carefully over the triangle.
\[ Cov(X,Y) = 4/9 - (4/5 \times 8/15) = 4/225 \]
Find the correlation coefficient for the runners in Problem 11.
You need \( E[X^2] \) and \( E[Y^2] \) to find variances.
\[ \rho = \frac{4/225}{\sqrt{2/75 \times 11/225}} = 4/\sqrt{66} \approx 0.492 \]
\( f(x,y) = 2, 0 < x < y < 1 \). Determine the correlation coefficient.
The total area is 1/2, so height \( C=2 \) is correct. The variables are dependent.
Calculation yields:
\[ \rho = \frac{1/36}{\sqrt{1/18 \times 1/18}} = 1/2 = 0.5 \]
Probabilities for 0, 1, 2, or 3 failures are 0.4, 0.3, 0.2, 0.1. Find mean and variance.
Purely discrete PMF. Sum \( x \cdot P(x) \) and \( x^2 \cdot P(x) \).
Mean: \( 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 1.0 \)
\( E[X^2] \): \( 0(0.4) + 1(0.3) + 4(0.2) + 9(0.1) = 2.0 \)
Variance: \( 2.0 - 1.0^2 = 1.0 \)
\( Y = 3X - 2 \), where \( X \) has density \( f(x) = \frac{1}{4} e^{-x/4}, x > 0 \). Find mean and variance of Y.
Identify X as an exponential distribution with parameter \(\lambda = 1/4\).
1. For \( X \sim Expo(1/4) \): \( E[X] = 4 \), \( Var(X) = 16 \).
2. \( E[Y] = 3(4) - 2 = 10 \).
3. \( Var(Y) = 3^2 \times 16 = 144 \).