Covariance - Bernoulli Distribution

Sorry, I forgot to include the 1/2 in the equation. In summary, we can find the covariance of random variables X and Y by using the formula Cov(X,Y) = E(XY) - (EX)(EY) and solving for E(XY) using the probabilities of X and the conditional expectations of Y. Alternatively, we can also use the formula cov(x,y) = \frac{p(p-1)}{2} if we have a uniform distribution for Y.
  • #1
Scootertaj
97
0
1. Consider the random variables X,Y where X~B(1,p) and
f(y|x=0) = 1/2 0<y<2
f(y|x=1) = 1 0<y<1

Find cov(x,y)




Homework Equations


Cov(x,y) = E(XY) - E(X)E(Y) = E[(x-E(x))(y-E(y))]
E(XY)=E[XE(Y|X)]



The Attempt at a Solution


E(X) = p (known since it's Bernoulli, can also be proven
E(Y) = [itex]\int Y*1/2 dy[/itex] 0 to 2 + [itex]\int Y*1[/itex] 0 to 1 = 3/2
I'm not sure E(Y) is right.

If this is right, I still don't know how to solve E(XY).

Could we do cov(x,y) =∫ ∫(x-p)(y-3/2)dxdy from 0 to 1, 0 to 2 ?

Thoughts?
 
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  • #2
Scootertaj said:
1. Consider the random variables X,Y where X~B(1,p) and
f(y|x=0) = 1/2 0<y<2
f(y|x=1) = 1 0<y<1

Find cov(x,y)




Homework Equations


Cov(x,y) = E(XY) - E(X)E(Y) = E[(x-E(x))(y-E(y))]
E(XY)=E[XE(Y|X)]



The Attempt at a Solution


E(X) = p (known since it's Bernoulli, can also be proven
E(Y) = [itex]\int Y*1/2 dy[/itex] 0 to 2 + [itex]\int Y*1[/itex] 0 to 1 = 3/2
I'm not sure E(Y) is right.
Your E(Y) is not correct. Rather than inputtting the coniditional distributions to start, try writing the fromula for E(Y) and work from that to see where the conditional distributions can be used.
Scootertaj said:
If this is right, I still don't know how to solve E(XY).

Could we do cov(x,y) =∫ ∫(x-p)(y-3/2)dxdy from 0 to 1, 0 to 2 ?

Thoughts?
 
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  • #3
Ya I thought so.
Well, E(Y) = ∫y*f(y)dy = ∫y*(∫f(x,y)dx)dy or = ∫y*f(x,y)*f(x|y)dy
But, I can't see how to use f(y|x=0) and f(y|x=1)

We do know that f(x) = px(1-p)1-x
 
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  • #4
though equivalent, the discrete veiw point for the probability mass function may be simpler to envisage here:
f(x) = p, if x=1
f(x) = (1-p), if x=0
f(x) = 0, otherwise

Now the expectation of a function of x, say g(x) will be:
[tex] E[g(x)] = \sum_{x_i} g(x_i)f(x_i) = pg(1)+(1-p)g(0)[/tex]

If x were continuously distributed, then the marginal distribution for Y is given by
[tex] f_Y(y) = \int f_{X,Y}(x,y)dx =\int f_{Y}(y|X=x)f_X(x)dx [/tex]

As x is a discrete variable, write teh marginal distribution of Y in terms of the discrete possibilties for x, and teh probabilites p? It may help to think of the integrand above as a function of x...
 
  • #5
lanedance said:
though equivalent, the discrete veiw point for the probability mass function may be simpler to envisage here:
f(x) = p, if x=1
f(x) = (1-p), if x=0
f(x) = 0, otherwise

Now the expectation of a function of x, say g(x) will be:
[tex] E[g(x)] = \sum_{x_i} g(x_i)f(x_i) = pg(1)+(1-p)g(0)[/tex]

If x were continuously distributed, then the marginal distribution for Y is given by
[tex] f_Y(y) = \int f_{X,Y}(x,y)dx =\int f_{Y}(y|X=x)f_X(x)dx [/tex]

As x is a discrete variable, write teh marginal distribution of Y in terms of the discrete possibilties for x, and teh probabilites p? It may help to think of the integrand above as a function of x...

So are you saying Y is g(x). But, we know f(y|x=0) and f(y|x=1) but do we know g(1), g(0)?
I tried thinking of it as a discrete, but couldn't we just use
[tex] f_Y(y) = \int f_{X,Y}(x,y)dx =\int f_{Y}(y|X=x)f_X(x)dx [/tex]
since we would get [tex] \int_0^2 1/2*1*(1-p)dx[/tex] + [tex] \int_0^1 (1*p)dx[/tex] since f(x) = px(1-p)1-x
Sorry, I'm just struggling to understand how to use the conditional pdf in this case.
 
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  • #6
Scootertaj said:
So are you saying Y is g(x). But, we know f(y|x=0) and f(y|x=1) but do we know g(1), g(0)?
I tried thinking of it as a discrete, but couldn't we just use
[tex] f_Y(y) = \int f_{X,Y}(x,y)dx =\int f_{Y}(y|X=x)f_X(x)dx [/tex]
since we would get [tex] \int_0^2 1/2*1*(1-p)dx[/tex] + [tex] \int_0^1 (1*p)dx[/tex] since f(x) = px(1-p)1-x
Sorry, I'm just struggling to understand how to use the conditional pdf in this case.

You cannot integrate over x because X iis a discrete random variable and so does not have a probability density.

You wrote the formula E(XY)=E[XE(Y|X)] in your original post. Do you understand what it MEANS? Can you write it out explicitly in terms of the possible values of X and their probabilities? Figuring out how to do that is Step 1 in the solution (or, at least, Step 1 in one approach to the solution).

RGV
 
  • #7
This is where I am now:

f(y|x=0) = 1/2 for 0<y<2 and f(y|x=1) = 1 0<y<1--> uniform distribution --> E(Y|x=0) = 1 ; E(Y|x=1) = 1/2
Also, E(Y) = E(Y|x=0)P(x=0) + E(Y|x=1)P(x=1) = 1-p/2
So, P(x=0,y) = (1-p)/3 for any y=0,1,2
P(x=1,y) = 1/2 for y=0,1 and 0 for y=2

Then,
[tex]cov(x,y) = \sum_{y=0}^2 \sum_{x=0}^1 (x-p)(y-(1-p/2))P(x,y) = \frac{p(p-1)}{2}[/tex]

Is this right?
 
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  • #8
Scootertaj said:
This is where I am now:

f(y|x=0) = 1/2 for 0<y<2 and f(y|x=1) = 1 0<y<1--> uniform distribution --> E(Y|x=0) = 1 ; E(Y|x=1) = 1/2
Also, E(Y) = E(Y|x=0)P(x=0) + E(Y|x=1)P(x=1) = 1-p/2
So, P(x=0,y) = (1-p)/3 for any y=0,1,2
P(x=1,y) = 1/2 for y=0,1 and 0 for y=2

Then,
[tex]cov(x,y) = \sum_{y=0}^2 \sum_{x=0}^1 (x-p)(y-(1-p/2))P(x,y) = \frac{p(p-1)}{2}[/tex]

Is this right?

Yes, it's OK. But, an easier way would be to compute E(XY) = P(X=0)*E(XY|X=0) + P(X=1)*E(XY|X=1) and to use Cov(X,Y) = E(XY) - (EX)(EY).

RGV
 
  • #9
Ray Vickson said:
Yes, it's OK. But, an easier way would be to compute E(XY) = P(X=0)*E(XY|X=0) + P(X=1)*E(XY|X=1) and to use Cov(X,Y) = E(XY) - (EX)(EY).

RGV
Hmm, I thought about that but when I was thinking about it it seemed difficult to calculate E(XY|X=0) for example. That is probably an oversight of mine, though.

But, [tex]cov(x,y) = \frac{p(p-1)}{2}[/tex] is correct?
 
  • #10
Scootertaj said:
Hmm, I thought about that but when I was thinking about it it seemed difficult to calculate E(XY|X=0) for example. That is probably an oversight of mine, though.

But, [tex]cov(x,y) = \frac{p(p-1)}{2}[/tex] is correct?

Easiest thing in the world: E(XY|X=0) = 0 (!) E(XY|X=1) = 1*E(Y|X=1).

I do not wish to answer the last question you asked.

RGV
 
  • #11
Doh, that was easy. It also makes a lot of sense. Both give the same answer so it's nice to see I did my summations correct.

Thanks a lot!
 
  • #12
I think I am being really stupid here but for the E(XY) part can you write down the full working I am really stuck on it, thanks
 
  • #13
What do you mean?
 
  • #14
Easiest thing in the world: E(XY|X=0) = 0 (!) E(XY|X=1) = 1*E(Y|X=1)

from this help that you got what do you do, what's the final answer?
 
  • #15
[tex]E(XY)=E(xy|x=0)P(x=0) + E(xy|x=1)P(x=1)[/tex]
 
  • #16
yep then i get that the original equation its what to do after, am i expecting an answer in terms of p and q? so i understand that the part E(XY|X=0) = 0 (!) and that E(XY|X=1) = 1*E(Y|X=1)
so does that mean I am left with E(XY)=E(Y|X=1) ?? its from here i am stuck, sorry if I am completely missing the point!
 
  • #17
Yes, you should get an answer in terms of p.
Just use the E(XY) formula above and recall the formula for cov(X,Y).
 
  • #18
Cov(XY)=E(XY) - E(X)E(Y) if I am not mistaken?
sorry I am being slow but i don't understand how you get E(Y|X=1) into terms of p and q
 
  • #19
cg7193 said:
Cov(XY)=E(XY) - E(X)E(Y) if I am not mistaken?
sorry I am being slow but i don't understand how you get E(Y|X=1) into terms of p and q
Yes, [tex]cov(X,Y) = E(XY)-E(X)E(Y)[/tex].
Moreover, [tex]E(XY) = E(XY|X=0)P(X=0) + E(XY|X=1)P(X=1)[/tex]
Now, find P(X=0), P(X=1) (this should be easy).
But, you are asking about how to find E(Y|X=1)? Well, we have a formula for f(Y|X=1) don't we? It's a horizontal line at 1 from 0 to 1. Then, the expected value (E(Y|X=1)) should be right in the middle.
Same idea for E(Y|X=0).
 
  • #20
ok ill give it a go now, thank you so much!
 

1. What is covariance?

Covariance is a measure of how two variables change together. It indicates the direction and strength of the relationship between two variables. A positive covariance means that the variables tend to increase or decrease together, while a negative covariance means that one variable tends to increase while the other decreases.

2. What is the Bernoulli distribution?

The Bernoulli distribution is a discrete probability distribution that models the outcomes of a single experiment with two possible outcomes - success and failure. It is often used to model random, binary events such as coin flips or yes/no scenarios.

3. How are covariance and the Bernoulli distribution related?

Covariance can be calculated for any two variables, including those that follow a Bernoulli distribution. In fact, the covariance between two Bernoulli-distributed variables is equal to the product of the probability of success for each variable, multiplied by the difference between those probabilities.

4. What does a high covariance between two variables indicate?

A high covariance between two variables indicates that they are strongly related and tend to change together. However, it does not necessarily mean that there is a causal relationship between the variables. It is important to also consider the magnitude of the covariance and the individual variances of each variable.

5. How is covariance used in data analysis?

Covariance is often used in data analysis to understand the relationship between two variables and to identify patterns or trends in the data. It can also help to identify potential predictors or factors that may influence an outcome. Additionally, covariance is a key component in calculating the correlation coefficient, which measures the strength and direction of a linear relationship between two variables.

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