Each day, Maude has a 1% chance of losing her cell phone (her behavior on different days is independent). Each day, Maude has a 3% chance of forgetting to eat breakfast (again, her behavior on different days is independent). Her breakfast and cell phone habits are independent. Let X be the number of days until she first loses her cell phone. Let Y be the number of days until she first forgets to eat breakfast. (Here, X and Y are independent). 1. Find the joint mass of X and Y. 2. Find the joint cdf of X and Y and briefly explain what FX,Y(x, y) represents in the context of the problem. 3. Find the conditional pmf pY|X(y|x).
Each day, Maude has a 1% chance of losing her cell phone (her behavior on different days is independent). Each day, Maude has a 3% chance of forgetting to eat breakfast (again, her behavior on different days is independent). Her breakfast and cell phone habits are independent. Let X be the number of days until she first loses her cell phone. Let Y be the number of days until she first forgets to eat breakfast. (Here, X and Y are independent). 1. Find the joint mass of X and Y. 2. Find the joint cdf of X and Y and briefly explain what FX,Y(x, y) represents in the context of the problem. 3. Find the conditional pmf pY|X(y|x).
The Correct Answer and Explanation is:
We are given:
- Probability Maude loses her phone on any day: pX=0.01p_X = 0.01
- Probability Maude forgets breakfast on any day: pY=0.03p_Y = 0.03
- XX: number of days until first lost phone (Geometric with parameter pXp_X)
- YY: number of days until first forgotten breakfast (Geometric with parameter pYp_Y)
- XX and YY are independent
1. Joint Mass Function pX,Y(x,y)p_{X,Y}(x,y)
Since X∼Geom(0.01)X \sim \text{Geom}(0.01) and Y∼Geom(0.03)Y \sim \text{Geom}(0.03), and they are independent: pX,Y(x,y)=P(X=x,Y=y)=P(X=x)⋅P(Y=y)p_{X,Y}(x, y) = P(X = x, Y = y) = P(X = x) \cdot P(Y = y) =(1−0.01)x−1(0.01)⋅(1−0.03)y−1(0.03)= (1 – 0.01)^{x-1}(0.01) \cdot (1 – 0.03)^{y-1}(0.03) =(0.99)x−1(0.01)⋅(0.97)y−1(0.03),x,y=1,2,3,…= (0.99)^{x-1}(0.01) \cdot (0.97)^{y-1}(0.03), \quad x, y = 1, 2, 3, \dots
2. Joint CDF FX,Y(x,y)=P(X≤x,Y≤y)F_{X,Y}(x,y) = P(X \leq x, Y \leq y)
Since XX and YY are independent: FX,Y(x,y)=P(X≤x)⋅P(Y≤y)F_{X,Y}(x, y) = P(X \leq x) \cdot P(Y \leq y)
Using the CDF of geometric distributions: P(X≤x)=1−(1−pX)x=1−(0.99)xP(X \leq x) = 1 – (1 – p_X)^x = 1 – (0.99)^x P(Y≤y)=1−(1−pY)y=1−(0.97)yP(Y \leq y) = 1 – (1 – p_Y)^y = 1 – (0.97)^y
So: FX,Y(x,y)=[1−(0.99)x]⋅[1−(0.97)y]F_{X,Y}(x, y) = \left[1 – (0.99)^x\right] \cdot \left[1 – (0.97)^y\right]
Interpretation:
FX,Y(x,y)F_{X,Y}(x, y) gives the probability that Maude loses her phone on or before day xx and forgets to eat breakfast on or before day yy.
3. Conditional PMF pY∣X(y∣x)p_{Y|X}(y|x)
By definition: pY∣X(y∣x)=pX,Y(x,y)P(X=x)p_{Y|X}(y|x) = \frac{p_{X,Y}(x,y)}{P(X = x)}
Since they are independent: pY∣X(y∣x)=P(X=x)⋅P(Y=y)P(X=x)=P(Y=y)p_{Y|X}(y|x) = \frac{P(X = x) \cdot P(Y = y)}{P(X = x)} = P(Y = y)
So: pY∣X(y∣x)=(0.97)y−1(0.03)p_{Y|X}(y|x) = (0.97)^{y-1}(0.03)
Interpretation:
Because XX and YY are independent, knowing the value of XX gives no information about YY, so the conditional distribution of YY given XX is just the marginal distribution of YY.
Summary Explanation
This problem models the time until Maude experiences two independent events: losing her phone and forgetting breakfast. These are both rare daily events, modeled using geometric distributions, where the variable represents the number of days until the first occurrence.
The joint PMF pX,Y(x,y)p_{X,Y}(x,y) is the product of the individual PMFs because the events are independent. The geometric PMF reflects that each day is a Bernoulli trial, and the event has not occurred on the previous days.
The joint CDF FX,Y(x,y)F_{X,Y}(x, y) gives the probability that both events happen within a certain number of days. Because of independence, we again multiply their individual CDFs. This CDF is useful for computing probabilities involving bounds, such as the chance that both events happen within a week.
The conditional PMF pY∣X(y∣x)p_{Y|X}(y|x) simplifies due to independence: learning that Maude lost her phone on day xx gives no information about when she might forget breakfast. Therefore, the conditional distribution of YY given XX is just the marginal distribution of YY.
This problem showcases how geometric distributions and independence combine in joint and conditional probability computations. The scenario is realistic, modeling rare, independent personal habits using standard discrete distributions.
