professor chooses student out of 23 girls and 17 boys then repeats 15 times, no replace. X = total # of boys....
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Hey Guys, so the above is a HW question I was recently assigned. I got it incorrect, and my professor posted the correct answers, as shown in the lower half of the image.
I quite easily got the answer for E[X] by simply doing (17/40)(15) = 6.375, which matches the answers from the image. I did this because it was repeated trials without replacement, so I just used indicator random variables, where I1 is the probability of 1 male being chosen, etc. Then, E[X] = E[I1] + E[I2] + ... + E[I15] = 15E[I1] = 15(17/40)
I used (17/40) for E[I1] because I1 can be 1 with 17/40 chance or 0 with 23/40 chance, so E[I1] = (17/40)(1) + (23/40)(0) = 17/40.
Now, for Variance I tried to do a similar thing, and say Var(X) = Var(I1) + Var(I2) ... Var(I15) = 15 * Var(I1)
to calculate Var(I1) I used the fact Var(x) = E[x^2] - (E[x]^2)
So Var(I1) = (17/40) - (17/40)^2 = 0.2444375
15Var(I1) = 3.67; this is not correct. I would appreciate any assistance in trying to figure out where I went wrong as I have an exam coming up on this material, and am very confused about what's wrong with my logic.
probability-theory probability-distributions random-variables variance expected-value
asked Nov 22 at 16:15
JackDremo
1
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up vote
-1
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favorite
Problem on Imgur
Hey Guys, so the above is a HW question I was recently assigned. I got it incorrect, and my professor posted the correct answers, as shown in the lower half of the image.
I quite easily got the answer for E[X] by simply doing (17/40)(15) = 6.375, which matches the answers from the image. I did this because it was repeated trials without replacement, so I just used indicator random variables, where I1 is the probability of 1 male being chosen, etc. Then, E[X] = E[I1] + E[I2] + ... + E[I15] = 15E[I1] = 15(17/40)
I used (17/40) for E[I1] because I1 can be 1 with 17/40 chance or 0 with 23/40 chance, so E[I1] = (17/40)(1) + (23/40)(0) = 17/40.
Now, for Variance I tried to do a similar thing, and say Var(X) = Var(I1) + Var(I2) ... Var(I15) = 15 * Var(I1)
to calculate Var(I1) I used the fact Var(x) = E[x^2] - (E[x]^2)
So Var(I1) = (17/40) - (17/40)^2 = 0.2444375
15Var(I1) = 3.67; this is not correct. I would appreciate any assistance in trying to figure out where I went wrong as I have an exam coming up on this material, and am very confused about what's wrong with my logic.
probability-theory probability-distributions random-variables variance expected-value
asked Nov 22 at 16:15
JackDremo
1
$(I_1+I_2+dots+I_{15})^2 neq (I_1^2+I_2^2+dots+I_{15}^2)$ and $Var(X)neq Var(I_1)+Var(I_2)+dots+Var(I_{15})$
– JMoravitz
Nov 22 at 16:19
I see, do you think you could perhaps point me in the right direction, because I have no clue where to go from here. I have read the Section of the textbook a few times, but I can't find the relevant information.
– JackDremo
Nov 22 at 16:21
$E[I_1]=E[I_1^2]$ is true however $X$ is not itself an indicator random variable and there is no reason to suspect that $E[X]$ is the same as $E[X^2]$ here, quite the opposite in fact we should expect them to be different.
– JMoravitz
Nov 22 at 16:24
As for how to continue, notice what $(I_1+I_2+dots+I_{15})^2$ is equal to. You have some terms like $I_k^2$ which simplify easily and you have other terms like $I_jI_k$ with $jneq k$. Notice that $I_jI_k=begin{cases}1&text{if both student j and k were selected at least once}\0&text{otherwise}end{cases}$ and you should be able to come up with a probability distribution for $I_jI_k$. Calculate $E[I_j]$ and calculate $E[I_jI_k]$, figure out how many terms are of the form $I_j$ and how many are of the form $I_jI_k$ and apply linearity of expectation.
– JMoravitz
Nov 22 at 16:26
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Problem on Imgur
Hey Guys, so the above is a HW question I was recently assigned. I got it incorrect, and my professor posted the correct answers, as shown in the lower half of the image.
I quite easily got the answer for E[X] by simply doing (17/40)(15) = 6.375, which matches the answers from the image. I did this because it was repeated trials without replacement, so I just used indicator random variables, where I1 is the probability of 1 male being chosen, etc. Then, E[X] = E[I1] + E[I2] + ... + E[I15] = 15E[I1] = 15(17/40)
I used (17/40) for E[I1] because I1 can be 1 with 17/40 chance or 0 with 23/40 chance, so E[I1] = (17/40)(1) + (23/40)(0) = 17/40.
Now, for Variance I tried to do a similar thing, and say Var(X) = Var(I1) + Var(I2) ... Var(I15) = 15 * Var(I1)
to calculate Var(I1) I used the fact Var(x) = E[x^2] - (E[x]^2)
So Var(I1) = (17/40) - (17/40)^2 = 0.2444375
15Var(I1) = 3.67; this is not correct. I would appreciate any assistance in trying to figure out where I went wrong as I have an exam coming up on this material, and am very confused about what's wrong with my logic.
probability-theory probability-distributions random-variables variance expected-value
asked Nov 22 at 16:15
JackDremo
1
Problem on Imgur
Hey Guys, so the above is a HW question I was recently assigned. I got it incorrect, and my professor posted the correct answers, as shown in the lower half of the image.
I quite easily got the answer for E[X] by simply doing (17/40)(15) = 6.375, which matches the answers from the image. I did this because it was repeated trials without replacement, so I just used indicator random variables, where I1 is the probability of 1 male being chosen, etc. Then, E[X] = E[I1] + E[I2] + ... + E[I15] = 15E[I1] = 15(17/40)
I used (17/40) for E[I1] because I1 can be 1 with 17/40 chance or 0 with 23/40 chance, so E[I1] = (17/40)(1) + (23/40)(0) = 17/40.
Now, for Variance I tried to do a similar thing, and say Var(X) = Var(I1) + Var(I2) ... Var(I15) = 15 * Var(I1)
to calculate Var(I1) I used the fact Var(x) = E[x^2] - (E[x]^2)
So Var(I1) = (17/40) - (17/40)^2 = 0.2444375
15Var(I1) = 3.67; this is not correct. I would appreciate any assistance in trying to figure out where I went wrong as I have an exam coming up on this material, and am very confused about what's wrong with my logic.
probability-theory probability-distributions random-variables variance expected-value
probability-theory probability-distributions random-variables variance expected-value
asked Nov 22 at 16:15
JackDremo
1
asked Nov 22 at 16:15
JackDremo
1
asked Nov 22 at 16:15
JackDremo
1
asked Nov 22 at 16:15
JackDremo
1
asked Nov 22 at 16:15
JackDremo
1
1
$(I_1+I_2+dots+I_{15})^2 neq (I_1^2+I_2^2+dots+I_{15}^2)$ and $Var(X)neq Var(I_1)+Var(I_2)+dots+Var(I_{15})$
– JMoravitz
Nov 22 at 16:19
I see, do you think you could perhaps point me in the right direction, because I have no clue where to go from here. I have read the Section of the textbook a few times, but I can't find the relevant information.
– JackDremo
Nov 22 at 16:21
$E[I_1]=E[I_1^2]$ is true however $X$ is not itself an indicator random variable and there is no reason to suspect that $E[X]$ is the same as $E[X^2]$ here, quite the opposite in fact we should expect them to be different.
– JMoravitz
Nov 22 at 16:24
As for how to continue, notice what $(I_1+I_2+dots+I_{15})^2$ is equal to. You have some terms like $I_k^2$ which simplify easily and you have other terms like $I_jI_k$ with $jneq k$. Notice that $I_jI_k=begin{cases}1&text{if both student j and k were selected at least once}\0&text{otherwise}end{cases}$ and you should be able to come up with a probability distribution for $I_jI_k$. Calculate $E[I_j]$ and calculate $E[I_jI_k]$, figure out how many terms are of the form $I_j$ and how many are of the form $I_jI_k$ and apply linearity of expectation.
– JMoravitz
Nov 22 at 16:26
add a comment |
$(I_1+I_2+dots+I_{15})^2 neq (I_1^2+I_2^2+dots+I_{15}^2)$ and $Var(X)neq Var(I_1)+Var(I_2)+dots+Var(I_{15})$
– JMoravitz
Nov 22 at 16:19
I see, do you think you could perhaps point me in the right direction, because I have no clue where to go from here. I have read the Section of the textbook a few times, but I can't find the relevant information.
– JackDremo
Nov 22 at 16:21
$E[I_1]=E[I_1^2]$ is true however $X$ is not itself an indicator random variable and there is no reason to suspect that $E[X]$ is the same as $E[X^2]$ here, quite the opposite in fact we should expect them to be different.
– JMoravitz
Nov 22 at 16:24
As for how to continue, notice what $(I_1+I_2+dots+I_{15})^2$ is equal to. You have some terms like $I_k^2$ which simplify easily and you have other terms like $I_jI_k$ with $jneq k$. Notice that $I_jI_k=begin{cases}1&text{if both student j and k were selected at least once}\0&text{otherwise}end{cases}$ and you should be able to come up with a probability distribution for $I_jI_k$. Calculate $E[I_j]$ and calculate $E[I_jI_k]$, figure out how many terms are of the form $I_j$ and how many are of the form $I_jI_k$ and apply linearity of expectation.
– JMoravitz
Nov 22 at 16:26
$(I_1+I_2+dots+I_{15})^2 neq (I_1^2+I_2^2+dots+I_{15}^2)$ and $Var(X)neq Var(I_1)+Var(I_2)+dots+Var(I_{15})$
– JMoravitz
Nov 22 at 16:19
$(I_1+I_2+dots+I_{15})^2 neq (I_1^2+I_2^2+dots+I_{15}^2)$ and $Var(X)neq Var(I_1)+Var(I_2)+dots+Var(I_{15})$
– JMoravitz
Nov 22 at 16:19
I see, do you think you could perhaps point me in the right direction, because I have no clue where to go from here. I have read the Section of the textbook a few times, but I can't find the relevant information.
– JackDremo
Nov 22 at 16:21
I see, do you think you could perhaps point me in the right direction, because I have no clue where to go from here. I have read the Section of the textbook a few times, but I can't find the relevant information.
– JackDremo
Nov 22 at 16:21
$E[I_1]=E[I_1^2]$ is true however $X$ is not itself an indicator random variable and there is no reason to suspect that $E[X]$ is the same as $E[X^2]$ here, quite the opposite in fact we should expect them to be different.
– JMoravitz
Nov 22 at 16:24
$E[I_1]=E[I_1^2]$ is true however $X$ is not itself an indicator random variable and there is no reason to suspect that $E[X]$ is the same as $E[X^2]$ here, quite the opposite in fact we should expect them to be different.
– JMoravitz
Nov 22 at 16:24
As for how to continue, notice what $(I_1+I_2+dots+I_{15})^2$ is equal to. You have some terms like $I_k^2$ which simplify easily and you have other terms like $I_jI_k$ with $jneq k$. Notice that $I_jI_k=begin{cases}1&text{if both student j and k were selected at least once}\0&text{otherwise}end{cases}$ and you should be able to come up with a probability distribution for $I_jI_k$. Calculate $E[I_j]$ and calculate $E[I_jI_k]$, figure out how many terms are of the form $I_j$ and how many are of the form $I_jI_k$ and apply linearity of expectation.
– JMoravitz
Nov 22 at 16:26
As for how to continue, notice what $(I_1+I_2+dots+I_{15})^2$ is equal to. You have some terms like $I_k^2$ which simplify easily and you have other terms like $I_jI_k$ with $jneq k$. Notice that $I_jI_k=begin{cases}1&text{if both student j and k were selected at least once}\0&text{otherwise}end{cases}$ and you should be able to come up with a probability distribution for $I_jI_k$. Calculate $E[I_j]$ and calculate $E[I_jI_k]$, figure out how many terms are of the form $I_j$ and how many are of the form $I_jI_k$ and apply linearity of expectation.
– JMoravitz
Nov 22 at 16:26
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$(I_1+I_2+dots+I_{15})^2 neq (I_1^2+I_2^2+dots+I_{15}^2)$ and $Var(X)neq Var(I_1)+Var(I_2)+dots+Var(I_{15})$
– JMoravitz
Nov 22 at 16:19
I see, do you think you could perhaps point me in the right direction, because I have no clue where to go from here. I have read the Section of the textbook a few times, but I can't find the relevant information.
– JackDremo
Nov 22 at 16:21
$E[I_1]=E[I_1^2]$ is true however $X$ is not itself an indicator random variable and there is no reason to suspect that $E[X]$ is the same as $E[X^2]$ here, quite the opposite in fact we should expect them to be different.
– JMoravitz
Nov 22 at 16:24
As for how to continue, notice what $(I_1+I_2+dots+I_{15})^2$ is equal to. You have some terms like $I_k^2$ which simplify easily and you have other terms like $I_jI_k$ with $jneq k$. Notice that $I_jI_k=begin{cases}1&text{if both student j and k were selected at least once}\0&text{otherwise}end{cases}$ and you should be able to come up with a probability distribution for $I_jI_k$. Calculate $E[I_j]$ and calculate $E[I_jI_k]$, figure out how many terms are of the form $I_j$ and how many are of the form $I_jI_k$ and apply linearity of expectation.
– JMoravitz
Nov 22 at 16:26