Computation of the expectation $E(2^{X-2})$ for $X$ negative binomial
$begingroup$
If I have a coin with $ 0.6$ probability of getting $ H $and I throw it until I get $ H $ for the second time. If $ Y $ is the number of $ T$ I get and $ 2^Y $ is my revenue. how do I calculate my revenue expectancy ?
If I set $ X $ as my total number of throws then $ Y = X-2 $
I know $ X sim NB(2,0.6) $ so $ E[X] = frac{2}{0.6} = frac{10}{3} $
$$ E[2^Y] = E[2^{x-2}]$$
but now I need some algebraic manipulation so I'll have something like $ E[aX + b] $ ?
I also tried different direction based on the probability function of negative binomial $ P{X = i} = { i-1 choose r-1 } (1-p)^{i-r} cdot p^r$ $$ E[2^Y] = E[2^{x-2}] = sum_{x} g(x)cdot p_x(x) = sum_{i=2}^infty 2^i cdot {i-1choose r-1} cdot 0.4^{i-2} cdot 0.6^2 = sum_{i=2}^infty 2^i cdot { i-1 choose 1 }cdot 0.4^{i-2} cdot 0.6^2 = sum_{i=2}^infty 2^i cdot(i-1) cdot 0.4^{i-2}cdot 0.6^2$$
Any suggestions?
probability-theory expected-value negative-binomial
edited Dec 9 '18 at 10:09
Did
247k23222458
asked Dec 9 '18 at 9:12
bm1125bm1125
64016
$endgroup$
add a comment |
$begingroup$
If I have a coin with $ 0.6$ probability of getting $ H $and I throw it until I get $ H $ for the second time. If $ Y $ is the number of $ T$ I get and $ 2^Y $ is my revenue. how do I calculate my revenue expectancy ?
If I set $ X $ as my total number of throws then $ Y = X-2 $
I know $ X sim NB(2,0.6) $ so $ E[X] = frac{2}{0.6} = frac{10}{3} $
$$ E[2^Y] = E[2^{x-2}]$$
but now I need some algebraic manipulation so I'll have something like $ E[aX + b] $ ?
I also tried different direction based on the probability function of negative binomial $ P{X = i} = { i-1 choose r-1 } (1-p)^{i-r} cdot p^r$ $$ E[2^Y] = E[2^{x-2}] = sum_{x} g(x)cdot p_x(x) = sum_{i=2}^infty 2^i cdot {i-1choose r-1} cdot 0.4^{i-2} cdot 0.6^2 = sum_{i=2}^infty 2^i cdot { i-1 choose 1 }cdot 0.4^{i-2} cdot 0.6^2 = sum_{i=2}^infty 2^i cdot(i-1) cdot 0.4^{i-2}cdot 0.6^2$$
Any suggestions?
probability-theory expected-value negative-binomial
edited Dec 9 '18 at 10:09
Did
247k23222458
asked Dec 9 '18 at 9:12
bm1125bm1125
64016
$endgroup$
1
$begingroup$
Here are two basic facts to enable you to conclude: $$sum_{i=2}^infty 2^i(i-1)(0.4)^{i-2}=4sum_{j=0}^infty(j+1)(0.8)^j$$ and, for every $|x|<1$, $$sum_{j=0}^infty(j+1)x^j=frac d{dx}sum_{k=0}^infty x^k=frac d{dx}frac1{1-x}=frac1{(1-x)^2}$$
$endgroup$
– Did
Dec 9 '18 at 10:07
$begingroup$
Thanks for the answer. Took me a while to understand what you did there in the first line algebrically. Could you explain the second line?I have this identity I know $$ sum_{i=0}^infty x^i = frac{1}{1-x} $$ so not sure how did you use it for $ sum_{j=0}^infty (j+1)x^j $ ? Does it mean the final answer should be just $ frac{1}{0.2^2} $ ?
$endgroup$
– bm1125
Dec 9 '18 at 10:33
$begingroup$
As is written, "what I did" is considering the derivative of the series you know. // It seems all this is rather pointing to a "final answer" equal to $4/(0.2)^2=100$.
$endgroup$
– Did
Dec 9 '18 at 10:39
1
$begingroup$
Two things: first, in your formula $2^i$ should read $2^{i-2}$; secondly, you forgot to take into account the factor $(0.6)^2$. Correcting the first mistake divides $100$ by $4$, correcting the second mistake multiplies it by $(0.6)^2$. And $(100div4)cdot(0.6)^2=$ ...
$endgroup$
– Did
Dec 9 '18 at 10:53
add a comment |
$begingroup$
If I have a coin with $ 0.6$ probability of getting $ H $and I throw it until I get $ H $ for the second time. If $ Y $ is the number of $ T$ I get and $ 2^Y $ is my revenue. how do I calculate my revenue expectancy ?
If I set $ X $ as my total number of throws then $ Y = X-2 $
I know $ X sim NB(2,0.6) $ so $ E[X] = frac{2}{0.6} = frac{10}{3} $
$$ E[2^Y] = E[2^{x-2}]$$
but now I need some algebraic manipulation so I'll have something like $ E[aX + b] $ ?
I also tried different direction based on the probability function of negative binomial $ P{X = i} = { i-1 choose r-1 } (1-p)^{i-r} cdot p^r$ $$ E[2^Y] = E[2^{x-2}] = sum_{x} g(x)cdot p_x(x) = sum_{i=2}^infty 2^i cdot {i-1choose r-1} cdot 0.4^{i-2} cdot 0.6^2 = sum_{i=2}^infty 2^i cdot { i-1 choose 1 }cdot 0.4^{i-2} cdot 0.6^2 = sum_{i=2}^infty 2^i cdot(i-1) cdot 0.4^{i-2}cdot 0.6^2$$
Any suggestions?
probability-theory expected-value negative-binomial
edited Dec 9 '18 at 10:09
Did
247k23222458
asked Dec 9 '18 at 9:12
bm1125bm1125
64016
$endgroup$
If I have a coin with $ 0.6$ probability of getting $ H $and I throw it until I get $ H $ for the second time. If $ Y $ is the number of $ T$ I get and $ 2^Y $ is my revenue. how do I calculate my revenue expectancy ?
If I set $ X $ as my total number of throws then $ Y = X-2 $
I know $ X sim NB(2,0.6) $ so $ E[X] = frac{2}{0.6} = frac{10}{3} $
$$ E[2^Y] = E[2^{x-2}]$$
but now I need some algebraic manipulation so I'll have something like $ E[aX + b] $ ?
I also tried different direction based on the probability function of negative binomial $ P{X = i} = { i-1 choose r-1 } (1-p)^{i-r} cdot p^r$ $$ E[2^Y] = E[2^{x-2}] = sum_{x} g(x)cdot p_x(x) = sum_{i=2}^infty 2^i cdot {i-1choose r-1} cdot 0.4^{i-2} cdot 0.6^2 = sum_{i=2}^infty 2^i cdot { i-1 choose 1 }cdot 0.4^{i-2} cdot 0.6^2 = sum_{i=2}^infty 2^i cdot(i-1) cdot 0.4^{i-2}cdot 0.6^2$$
Any suggestions?
probability-theory expected-value negative-binomial
probability-theory expected-value negative-binomial
edited Dec 9 '18 at 10:09
Did
247k23222458
asked Dec 9 '18 at 9:12
bm1125bm1125
64016
edited Dec 9 '18 at 10:09
Did
247k23222458
asked Dec 9 '18 at 9:12
bm1125bm1125
64016
edited Dec 9 '18 at 10:09
Did
247k23222458
edited Dec 9 '18 at 10:09
Did
247k23222458
edited Dec 9 '18 at 10:09
Did
247k23222458
247k23222458
asked Dec 9 '18 at 9:12
bm1125bm1125
64016
asked Dec 9 '18 at 9:12
bm1125bm1125
64016
asked Dec 9 '18 at 9:12
bm1125bm1125
64016
64016
1
$begingroup$
Here are two basic facts to enable you to conclude: $$sum_{i=2}^infty 2^i(i-1)(0.4)^{i-2}=4sum_{j=0}^infty(j+1)(0.8)^j$$ and, for every $|x|<1$, $$sum_{j=0}^infty(j+1)x^j=frac d{dx}sum_{k=0}^infty x^k=frac d{dx}frac1{1-x}=frac1{(1-x)^2}$$
$endgroup$
– Did
Dec 9 '18 at 10:07
$begingroup$
Thanks for the answer. Took me a while to understand what you did there in the first line algebrically. Could you explain the second line?I have this identity I know $$ sum_{i=0}^infty x^i = frac{1}{1-x} $$ so not sure how did you use it for $ sum_{j=0}^infty (j+1)x^j $ ? Does it mean the final answer should be just $ frac{1}{0.2^2} $ ?
$endgroup$
– bm1125
Dec 9 '18 at 10:33
$begingroup$
As is written, "what I did" is considering the derivative of the series you know. // It seems all this is rather pointing to a "final answer" equal to $4/(0.2)^2=100$.
$endgroup$
– Did
Dec 9 '18 at 10:39
1
$begingroup$
Two things: first, in your formula $2^i$ should read $2^{i-2}$; secondly, you forgot to take into account the factor $(0.6)^2$. Correcting the first mistake divides $100$ by $4$, correcting the second mistake multiplies it by $(0.6)^2$. And $(100div4)cdot(0.6)^2=$ ...
$endgroup$
– Did
Dec 9 '18 at 10:53
add a comment |
1
$begingroup$
Here are two basic facts to enable you to conclude: $$sum_{i=2}^infty 2^i(i-1)(0.4)^{i-2}=4sum_{j=0}^infty(j+1)(0.8)^j$$ and, for every $|x|<1$, $$sum_{j=0}^infty(j+1)x^j=frac d{dx}sum_{k=0}^infty x^k=frac d{dx}frac1{1-x}=frac1{(1-x)^2}$$
$endgroup$
– Did
Dec 9 '18 at 10:07
$begingroup$
Thanks for the answer. Took me a while to understand what you did there in the first line algebrically. Could you explain the second line?I have this identity I know $$ sum_{i=0}^infty x^i = frac{1}{1-x} $$ so not sure how did you use it for $ sum_{j=0}^infty (j+1)x^j $ ? Does it mean the final answer should be just $ frac{1}{0.2^2} $ ?
$endgroup$
– bm1125
Dec 9 '18 at 10:33
$begingroup$
As is written, "what I did" is considering the derivative of the series you know. // It seems all this is rather pointing to a "final answer" equal to $4/(0.2)^2=100$.
$endgroup$
– Did
Dec 9 '18 at 10:39
1
$begingroup$
Two things: first, in your formula $2^i$ should read $2^{i-2}$; secondly, you forgot to take into account the factor $(0.6)^2$. Correcting the first mistake divides $100$ by $4$, correcting the second mistake multiplies it by $(0.6)^2$. And $(100div4)cdot(0.6)^2=$ ...
$endgroup$
– Did
Dec 9 '18 at 10:53
1
1
$begingroup$
Here are two basic facts to enable you to conclude: $$sum_{i=2}^infty 2^i(i-1)(0.4)^{i-2}=4sum_{j=0}^infty(j+1)(0.8)^j$$ and, for every $|x|<1$, $$sum_{j=0}^infty(j+1)x^j=frac d{dx}sum_{k=0}^infty x^k=frac d{dx}frac1{1-x}=frac1{(1-x)^2}$$
$endgroup$
– Did
Dec 9 '18 at 10:07
$begingroup$
Here are two basic facts to enable you to conclude: $$sum_{i=2}^infty 2^i(i-1)(0.4)^{i-2}=4sum_{j=0}^infty(j+1)(0.8)^j$$ and, for every $|x|<1$, $$sum_{j=0}^infty(j+1)x^j=frac d{dx}sum_{k=0}^infty x^k=frac d{dx}frac1{1-x}=frac1{(1-x)^2}$$
$endgroup$
– Did
Dec 9 '18 at 10:07
$begingroup$
Thanks for the answer. Took me a while to understand what you did there in the first line algebrically. Could you explain the second line?I have this identity I know $$ sum_{i=0}^infty x^i = frac{1}{1-x} $$ so not sure how did you use it for $ sum_{j=0}^infty (j+1)x^j $ ? Does it mean the final answer should be just $ frac{1}{0.2^2} $ ?
$endgroup$
– bm1125
Dec 9 '18 at 10:33
$begingroup$
Thanks for the answer. Took me a while to understand what you did there in the first line algebrically. Could you explain the second line?I have this identity I know $$ sum_{i=0}^infty x^i = frac{1}{1-x} $$ so not sure how did you use it for $ sum_{j=0}^infty (j+1)x^j $ ? Does it mean the final answer should be just $ frac{1}{0.2^2} $ ?
$endgroup$
– bm1125
Dec 9 '18 at 10:33
$begingroup$
As is written, "what I did" is considering the derivative of the series you know. // It seems all this is rather pointing to a "final answer" equal to $4/(0.2)^2=100$.
$endgroup$
– Did
Dec 9 '18 at 10:39
$begingroup$
As is written, "what I did" is considering the derivative of the series you know. // It seems all this is rather pointing to a "final answer" equal to $4/(0.2)^2=100$.
$endgroup$
– Did
Dec 9 '18 at 10:39
1
1
$begingroup$
Two things: first, in your formula $2^i$ should read $2^{i-2}$; secondly, you forgot to take into account the factor $(0.6)^2$. Correcting the first mistake divides $100$ by $4$, correcting the second mistake multiplies it by $(0.6)^2$. And $(100div4)cdot(0.6)^2=$ ...
$endgroup$
– Did
Dec 9 '18 at 10:53
$begingroup$
Two things: first, in your formula $2^i$ should read $2^{i-2}$; secondly, you forgot to take into account the factor $(0.6)^2$. Correcting the first mistake divides $100$ by $4$, correcting the second mistake multiplies it by $(0.6)^2$. And $(100div4)cdot(0.6)^2=$ ...
$endgroup$
– Did
Dec 9 '18 at 10:53
add a comment |
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$begingroup$
Here are two basic facts to enable you to conclude: $$sum_{i=2}^infty 2^i(i-1)(0.4)^{i-2}=4sum_{j=0}^infty(j+1)(0.8)^j$$ and, for every $|x|<1$, $$sum_{j=0}^infty(j+1)x^j=frac d{dx}sum_{k=0}^infty x^k=frac d{dx}frac1{1-x}=frac1{(1-x)^2}$$
$endgroup$
– Did
Dec 9 '18 at 10:07
$begingroup$
Thanks for the answer. Took me a while to understand what you did there in the first line algebrically. Could you explain the second line?I have this identity I know $$ sum_{i=0}^infty x^i = frac{1}{1-x} $$ so not sure how did you use it for $ sum_{j=0}^infty (j+1)x^j $ ? Does it mean the final answer should be just $ frac{1}{0.2^2} $ ?
$endgroup$
– bm1125
Dec 9 '18 at 10:33
$begingroup$
As is written, "what I did" is considering the derivative of the series you know. // It seems all this is rather pointing to a "final answer" equal to $4/(0.2)^2=100$.
$endgroup$
– Did
Dec 9 '18 at 10:39
1
$begingroup$
Two things: first, in your formula $2^i$ should read $2^{i-2}$; secondly, you forgot to take into account the factor $(0.6)^2$. Correcting the first mistake divides $100$ by $4$, correcting the second mistake multiplies it by $(0.6)^2$. And $(100div4)cdot(0.6)^2=$ ...
$endgroup$
– Did
Dec 9 '18 at 10:53