GATE DS&AI 2024 | Question: 55

Question

Two fair coins are tossed independently. $X$ is a random variable that takes a value of $1$ if both tosses are heads and $0$ otherwise. $Y$ is a random variable that takes a value of $1$ if at least one of the tosses is heads and $0$ otherwise.

The value of the covariance of $X$ and $Y$ is $\_\_\_\_\_\_\_$ (rounded off to three decimal places).

Answer 1

 

 

So,

 

$\mathbb{E}[X]=0*\frac{3}{4}+1*\frac{1}{4}=\frac{1}{4}$

 

$\mathbb{E}[Y]=0*\frac{1}{4}+1*\frac{3}{4}=\frac{3}{4}$

 

$\mathbb{E}[XY]=1*1*\frac{1}{4}=\frac{1}{4}$

 

$Cov[X,Y]=\mathbb{E}[XY]-\mathbb{E}[X]E[Y]=\frac{1}{16}$

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If we have a real-valued function $g(X_1,X_2)$ such that $g: R^2 \rightarrow R$ then in case of discrete distribution:

$E[g(X_1,X_2)]=\Sigma_{x_1} \Sigma_{x_2}g(X_1,X_2)\;  p_{X_1,X_2} (x_1,x_2)$

Where  $p_{X_1,X_2} (x_1,x_2)$ represents the joint probability distribution and $x_1$ and $x_2$ are instances of $X_1$ and $X_2.$

So, here, we need to find $E[XY],$ It means $g(X_1,X_2)=XY$

Hence,

$E[XY]=\Sigma_{x} \Sigma_{y}XY\;  p_{X,Y} (x,y)$ 

Where x and y are instances of $X$ and $Y.$

 and Joint Distribution is given in the above table as:

1. When $x=0,y=0$ then  $p_{X,Y} (x,y) =\frac{1}{4}$

2. When $x=0,y=1$ then  $p_{X,Y} (x,y) =\frac{1}{2}$

3. When $x=1,y=0$ then  $p_{X,Y} (x,y) =0$

4. When $x=1,y=1$ then  $p_{X,Y} (x,y) =\frac{1}{4}$

Therefore,

$E[XY]=0*0*\frac{1}{4}+0*1*\frac{1}{2}+1*0*0+1*1*\frac{1}{4}=\frac{1}{4}$

Comments

Can you eloborate more on calculating E[XY]?

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added the explanation for E[XY]. Let me know if you have any doubt.