Independent solution

How to solve this Joint Distributions question

Setup

Setup

Read the two Bernoulli success probabilities and the joint-success probability from the joint distribution.

Pr⁡(X=1)=0.05+0.10=0.15\Pr(X=1)=0.05+0.10=0.15
Pr⁡(Y=1)=0.45+0.10=0.55\Pr(Y=1)=0.45+0.10=0.55
E[XY]=Pr⁡(X=1,Y=1)=0.10\mathbb{E}[XY]=\Pr(X=1,Y=1)=0.10

Model

Model

For a Bernoulli variable, the variance is p(1-p). Center the product moment to obtain the covariance.

Var⁡(X)=0.15(0.85)=0.1275\operatorname{Var}(X)=0.15(0.85)=0.1275
Var⁡(Y)=0.55(0.45)=0.2475\operatorname{Var}(Y)=0.55(0.45)=0.2475
Cov⁡(X,Y)=0.10−(0.15)(0.55)=0.0175\operatorname{Cov}(X,Y)=0.10-(0.15)(0.55)=0.0175

Compute

Compute

Divide the covariance by the product of the two marginal standard deviations.

ρX,Y=0.0175(0.1275)(0.2475)\rho_{X,Y}=\frac{0.0175}{\sqrt{(0.1275)(0.2475)}}
ρX,Y=0.0985134105…\rho_{X,Y}=0.0985134105\ldots

Answer

Answer

The correlation rounds to three decimals as 0.099.

ρX,Y≈0.099(B)\boxed{\rho_{X,Y}\approx 0.099\quad\text{(B)}}