mean and variance: Let X be a discrete random variable such that X = 0 with probability 0.5 and X = 1 with probability 0.5.

[Meera007]

Let X be a discrete random variable such that X = 0 with probability 0.5 and X = 1 with probability 0.5. Let Y be a discrete random variable such that Y = 0 when X = 1 and Y = 1 when X = 0. What is the mean and variance of X + Y?

Hint: You can define W=X+Y and apply the expected value and variance formulas for a sum of two random variables.

Pls help to do this

 

[stapel]

Let X be a discrete random variable such that X = 0 with probability 0.5 and X = 1 with probability 0.5. Let Y be a discrete random variable such that Y = 0 when X = 1 and Y = 1 when X = 0. What is the mean and variance of X + Y?

Hint: You can define W=X+Y and apply the expected value and variance formulas for a sum of two random variables.

Pls help to do this

We'll be glad to help! But first we need to see what you've tried and where things are going wrong. ("Read Before Posting")

Please be complete. Thank you!

 

[Meera007]

Z = X +Y

mean = E(Z)=0+0.5+2⋅0.5 = 1.5
E(Z2)=0+0.5+4⋅0.5 = 2.5
Var(Z)=2.5−(1.5)2 = 0.25

is this correct ?

 

[BigBeachBanana]

Let X be a discrete random variable such that X = 0 with probability 0.5 and X = 1 with probability 0.5. Let Y be a discrete random variable such that Y = 0 when X = 1 and Y = 1 when X = 0. What is the mean and variance of X + Y?

Hint: You can define W=X+Y and apply the expected value and variance formulas for a sum of two random variables.

Pls help to do this

Start by finding the mean and variance of X and Y.

 

[Meera007]

Z = X +Y

mean = E(Z)=0+0.5+2⋅0.5 = 1.5
E(Z2)=0+0.5+4⋅0.5 = 2.5
Var(Z)=2.5−(1.5)2 = 0.25

mean 0.5 + .5 = 1

Variance X = .5*(0-.5) *(0-.5) + .5*(0-.5) *(0-.5) = .25
Variance Y= .5*(0-.5) *(0-.5) + .5*(0-.5) *(0-.5) = .25
X+Y = .25 +.25 = .5

So mean 1 and variance .5

Am I correct?

Thanks in advance

 

[blamocur]

Let X be a discrete random variable such that X = 0 with probability 0.5 and X = 1 with probability 0.5. Let Y be a discrete random variable such that Y = 0 when X = 1 and Y = 1 when X = 0. What is the mean and variance of X + Y?

Hint: You can define W=X+Y and apply the expected value and variance formulas for a sum of two random variables.

Pls help to do this

Looks to me that [imath]X+Y=1[/imath]. Am I missing something here?

 

[BigBeachBanana]

Variance X = .5*(0-.5) *(0-.5) + .5*(0-.5) *(0-.5) = .25
Variance Y= .5*(0-.5) *(0-.5) + .5*(0-.5) *(0-.5) = .25
X+Y = .25 +.25 = .5

So mean 1 and variance .5

Am I correct?

Thanks in advance

Variance is incorrect. Here's my suggestion. Find the distribution of [imath]W[/imath].

Notice:
If [imath]X=1[/imath], then [imath]X+Y =1[/imath]
If [imath]X=0[/imath], then [imath]X+Y=1[/imath]
Or [imath]\Pr(X+Y=1)=1[/imath]

With this information, we can construct the distribution for [imath]W[/imath].

[math]W= \begin{cases} 1 \quad \text{with prob. of } 1\\ 0 \quad \text{with prob. of } 0\\ \end{cases}[/math]What is [imath]E(W)?[/imath]
What is [imath]E(W^2)?[/imath]
What is [imath]Var(W) = E(W^2)-[E(W)]^2[/imath]

 

[BigBeachBanana]

Variance is incorrect. Here's my suggestion. Find the distribution of [imath]W[/imath].

Notice:
If [imath]X=1[/imath], then [imath]X+Y =1[/imath]
If [imath]X=0[/imath], then [imath]X+Y=1[/imath]
Or [imath]\Pr(X+Y=1)=1[/imath]

With this information, we can construct the distribution for [imath]W[/imath].

[math]W= \begin{cases} 1 \quad \text{with prob. of } 1\\ 0 \quad \text{with prob. of } 0\\ \end{cases}[/math]What is [imath]E(W)?[/imath]
What is [imath]E(W^2)?[/imath]
What is [imath]Var(W) = E(W^2)-[E(W)]^2[/imath]

Typo above:
[math]W= \begin{cases} 1 \quad \text{with prob. of } 1\\ \red{\text{else}} \quad \text{with prob. of } 0\\ \end{cases}[/math]