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Probability 1. Consider the random variable XX with the probability mass funct...
Jun 14, 2024
Solution by Steps
step 1
The moment generating function (MGF) of a random variable XX is defined as MX(t)=E[etX]M_X(t) = E[e^{tX}]
step 2
For the given PMF pX(x)=p(1−p)x−1p_X(x) = p(1-p)^{x-1}, we need to find E[etX]E[e^{tX}]
step 3
MX(t)=∑x=1∞etxpX(x)=∑x=1∞etxp(1−p)x−1M_X(t) = \sum_{x=1}^{\infty} e^{tx} p_X(x) = \sum_{x=1}^{\infty} e^{tx} p(1-p)^{x-1}
step 4
Simplifying, MX(t)=p∑x=1∞(et(1−p))x−1M_X(t) = p \sum_{x=1}^{\infty} (e^t (1-p))^{x-1}
step 5
This is a geometric series with the first term a=1a = 1 and common ratio r=et(1−p)r = e^t (1-p)
step 6
The sum of an infinite geometric series is a1−r\frac{a}{1-r}, so MX(t)=p1−et(1−p)M_X(t) = \frac{p}{1 - e^t (1-p)}
step 7
Therefore, the MGF of XX is MX(t)=pet1−(1−p)etM_X(t) = \frac{p e^t}{1 - (1-p) e^t}
# Part (b)
step 1
The mean of XX can be found using the MGF by differentiating MX(t)M_X(t) with respect to tt and evaluating at t=0t=0
step 2
MX′(t)=ddt(pet1−(1−p)et)M_X'(t) = \frac{d}{dt} \left( \frac{p e^t}{1 - (1-p) e^t} \right)
step 3
Using the quotient rule, MX′(t)=pet(1−(1−p)et)−pet(1−p)et(1−(1−p)et)2M_X'(t) = \frac{p e^t (1 - (1-p) e^t) - p e^t (1-p) e^t}{(1 - (1-p) e^t)^2}
step 4
Simplifying, MX′(t)=pet(1−(1−p)et−(1−p)et)(1−(1−p)et)2=pet(1−et)(1−(1−p)et)2M_X'(t) = \frac{p e^t (1 - (1-p) e^t - (1-p) e^t)}{(1 - (1-p) e^t)^2} = \frac{p e^t (1 - e^t)}{(1 - (1-p) e^t)^2}
step 5
Evaluating at t=0t=0, MX′(0)=p(1−1)(1−(1−p))2=p(1−1)p2=0p2=0M_X'(0) = \frac{p (1 - 1)}{(1 - (1-p))^2} = \frac{p (1 - 1)}{p^2} = \frac{0}{p^2} = 0
step 6
Therefore, the mean of XX is E[X]=MX′(0)=0E[X] = M_X'(0) = 0
Answer
The MGF of XX is pet1−(1−p)et\frac{p e^t}{1 - (1-p) e^t} and the mean of XX is 00.
Key Concept
Moment Generating Function (MGF)
Explanation
The MGF is used to find moments of a random variable, and the mean can be found by differentiating the MGF and evaluating at t=0t=0.
Question 2 # Part (a)
step 1
Given X∼N(μ=3,σ2=9)X \sim N(\mu=3, \sigma^2=9), we need to find P(z < X < 5)
step 2
Standardize XX to ZZ using Z=X−μσZ = \frac{X - \mu}{\sigma}
step 3
For X=zX = z, Z=z−33Z = \frac{z - 3}{3} and for X=5X = 5, Z=5−33=23Z = \frac{5 - 3}{3} = \frac{2}{3}
step 4
Therefore, P(z < X < 5) = P\left(\frac{z - 3}{3} < Z < \frac{2}{3}\right)
step 5
Using standard normal distribution tables, find the probabilities corresponding to Z=z−33Z = \frac{z - 3}{3} and Z=23Z = \frac{2}{3}
step 6
P(Z < \frac{2}{3}) \approx 0.7486 and P(Z < \frac{z - 3}{3}) depends on zz
step 7
Therefore, P(z < X < 5) = 0.7486 - P(Z < \frac{z - 3}{3})
# Part (b)
step 1
Given X∼N(μ=3,σ2=9)X \sim N(\mu=3, \sigma^2=9), we need to find P(X > 6)
step 2
Standardize XX to ZZ using Z=X−μσZ = \frac{X - \mu}{\sigma}
step 3
For X=6X = 6, Z=6−33=1Z = \frac{6 - 3}{3} = 1
step 4
Therefore, P(X > 6) = P\left(Z > 1\right)
step 5
Using standard normal distribution tables, P(Z > 1) = 1 - P(Z < 1) \approx 1 - 0.8413 = 0.1587
Answer
(a) P(z < X < 5) = 0.7486 - P(Z < \frac{z - 3}{3}) and (b) P(X > 6) = 0.1587.
Key Concept
Standard Normal Distribution
Explanation
Standardizing a normal random variable allows us to use standard normal distribution tables to find probabilities.
Question 3 # Part (a)
step 1
Given the PDF f(x)=C(4x−2x2)f(x) = C(4x - 2x^2) for 0 < x < 2, we need to find the value of CC
step 2
The total probability must equal 1, so ∫02C(4x−2x2) dx=1\int_{0}^{2} C(4x - 2x^2) \, dx = 1
step 3
Integrate: ∫02C(4x−2x2) dx=C[2x2−2x33]02\int_{0}^{2} C(4x - 2x^2) \, dx = C \left[ 2x^2 - \frac{2x^3}{3} \right]_{0}^{2}
step 4
Evaluate the integral: C[2(2)2−2(2)33]=C[8−163]=C[243−163]=C[83]C \left[ 2(2)^2 - \frac{2(2)^3}{3} \right] = C \left[ 8 - \frac{16}{3} \right] = C \left[ \frac{24}{3} - \frac{16}{3} \right] = C \left[ \frac{8}{3} \right]
step 5
Set the integral equal to 1: C[83]=1⇒C=38C \left[ \frac{8}{3} \right] = 1 \Rightarrow C = \frac{3}{8}
# Part (b)
step 1
We need to find P(X > 1)
step 2
P(X > 1) = \int_{1}^{2} \frac{3}{8} (4x - 2x^2) \, dx
step 3
Integrate: ∫1238(4x−2x2) dx=38[2x2−2x33]12\int_{1}^{2} \frac{3}{8} (4x - 2x^2) \, dx = \frac{3}{8} \left[ 2x^2 - \frac{2x^3}{3} \right]_{1}^{2}
step 4
Evaluate the integral: 38[2(2)2−2(2)33−(2(1)2−2(1)33)]=38[8−163−(2−23)]\frac{3}{8} \left[ 2(2)^2 - \frac{2(2)^3}{3} - (2(1)^2 - \frac{2(1)^3}{3}) \right] = \frac{3}{8} \left[ 8 - \frac{16}{3} - (2 - \frac{2}{3}) \right]
step 5
Simplify: 38[243−163−63+23]=38[43]=12\frac{3}{8} \left[ \frac{24}{3} - \frac{16}{3} - \frac{6}{3} + \frac{2}{3} \right] = \frac{3}{8} \left[ \frac{4}{3} \right] = \frac{1}{2}
Answer
(a) C=38C = \frac{3}{8} and (b) P(X > 1) = \frac{1}{2}.
Key Concept
Probability Density Function (PDF)
Explanation
The total probability under a PDF must equal 1, and probabilities can be found by integrating the PDF over the desired range.
Question 4 # Part (a)
step 1
Given X∼U(0,1)X \sim U(0,1) and Y=XnY = X^n, we need to find the CDF FY(y)F_Y(y)
step 2
FY(y)=P(Y≤y)=P(Xn≤y)F_Y(y) = P(Y \leq y) = P(X^n \leq y)
step 3
Since X∼U(0,1)X \sim U(0,1), P(X≤y1/n)=y1/nP(X \leq y^{1/n}) = y^{1/n} for 0≤y≤10 \leq y \leq 1
step 4
Therefore, FY(y)=y1/nF_Y(y) = y^{1/n} for 0≤y≤10 \leq y \leq 1
# Part (b)
step 1
To find the PDF fY(y)f_Y(y), differentiate the CDF FY(y)F_Y(y)
step 2
fY(y)=ddyFY(y)=ddyy1/n=1ny1/n−1f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} y^{1/n} = \frac{1}{n} y^{1/n - 1}
step 3
Therefore, fY(y)=1ny1/n−1f_Y(y) = \frac{1}{n} y^{1/n - 1} for 0≤y≤10 \leq y \leq 1
Answer
(a) FY(y)=y1/nF_Y(y) = y^{1/n} and (b) fY(y)=1ny1/n−1f_Y(y) = \frac{1}{n} y^{1/n - 1}.
Key Concept
Cumulative Distribution Function (CDF) and Probability Density Function (PDF)
Explanation
The CDF is the integral of the PDF, and the PDF is the derivative of the CDF.
Question 5 # Part (a)
step 1
Given the joint PDF fX,Y(x,y)=x+yf_{X,Y}(x,y) = x + y for 0 < x < 1 and 0 < y < 1, we need to find E[XY]E[XY], E[X]E[X], and E[Y]E[Y]
step 2
E[X]=∫01∫01x(x+y) dy dxE[X] = \int_{0}^{1} \int_{0}^{1} x (x + y) \, dy \, dx
step 3
Integrate with respect to yy: ∫01x[y22+xy]01 dx=∫01x[12+x] dx\int_{0}^{1} x \left[ \frac{y^2}{2} + xy \right]_{0}^{1} \, dx = \int_{0}^{1} x \left[ \frac{1}{2} + x \right] \, dx
step 4
Integrate with respect to xx: ∫01[x2+x2] dx=[x24+x33]01=14+13=712\int_{0}^{1} \left[ \frac{x}{2} + x^2 \right] \, dx = \left[ \frac{x^2}{4} + \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}
step 5
Similarly, E[Y]=∫01∫01y(x+y) dx dyE[Y] = \int_{0}^{1} \int_{0}^{1} y (x + y) \, dx \, dy
step 6
Integrate with respect to xx: ∫01y[x22+xy]01 dy=∫01y[12+y] dy\int_{0}^{1} y \left[ \frac{x^2}{2} + xy \right]_{0}^{1} \, dy = \int_{0}^{1} y \left[ \frac{1}{2} + y \right] \, dy
step 7
Integrate with respect to yy: ∫01[y2+y2] dy=[y24+y33]01=14+13=712\int_{0}^{1} \left[ \frac{y}{2} + y^2 \right] \, dy = \left[ \frac{y^2}{4} + \frac{y^3}{3} \right]_{0}^{1} = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}
step 8
E[XY]=∫01∫01xy(x+y) dy dxE[XY] = \int_{0}^{1} \int_{0}^{1} xy (x + y) \, dy \, dx
step 9
Integrate with respect to yy: ∫01x[y22+xy]01 dx=∫01x[12+x] dx\int_{0}^{1} x \left[ \frac{y^2}{2} + xy \right]_{0}^{1} \, dx = \int_{0}^{1} x \left[ \frac{1}{2} + x \right] \, dx
step 10
Integrate with respect to xx: ∫01[x2+x2] dx=[x24+x33]01=14+13=712\int_{0}^{1} \left[ \frac{x}{2} + x^2 \right] \, dx = \left[ \frac{x^2}{4} + \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}
# Part (b)
step 1
The covariance of XX and YY is given by Cov(X,Y)=E[XY]−E[X]E[Y]\text{Cov}(X,Y) = E[XY] - E[X]E[Y]
step 2
Using the values found: Cov(X,Y)=712−(712)2\text{Cov}(X,Y) = \frac{7}{12} - \left(\frac{7}{12}\right)^2
step 3
Simplify: Cov(X,Y)=712−49144=84144−49144=35144\text{Cov}(X,Y) = \frac{7}{12} - \frac{49}{144} = \frac{84}{144} - \frac{49}{144} = \frac{35}{144}
Answer
(a) E[X]=712E[X] = \frac{7}{12}, E[Y]=712E[Y] = \frac{7}{12}, E[XY]=712E[XY] = \frac{7}{12} and (b) Cov(X,Y)=35144\text{Cov}(X,Y) = \frac{35}{144}.
Key Concept
Expectation and Covariance
Explanation
Expectation is the mean value of a random variable, and covariance measures the joint variability of two random variables.
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