Mixing
On calculating conditional expectations
$begingroup$
Attempt:
Notice we can write $T = max (T_1, T_2)$. One can find the pmf:
$$ P(T leq t) = P(max(T_1,T_2) leq t) = P(T_1 leq t, T_2 leq t) = (1-e^{-t/2})^2$$
therefore $f_T(t) = (1-e^{t/2}) e^{-t/2} = e^{-t/2} - e^{-t}$
Now, $f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)} = dfrac{e^{-t/2} - e^{-t}}{e^{-1}} $
Thus, the expectation is
$$ E(T | T_1=1) = int t dfrac{e^{-t/2} - e^{-t}}{2e^{-1/2}} dt = e (2 - 1) = boxed{2e^{1/2}}$$
Now, for $b$, we evaluate $int_1^{infty} f_{T_1} (t) = 1 - P(1) = e^{-1/2} $
So, $E(T | T_1 > 1) = e^{1/2} $
Now, Is my approach correct so far? For part c), what would be the best approach? or just brute furce ?
probability-theory conditional-probability
edited Jan 27 at 10:34
Did
248k23226466
asked Jan 27 at 10:21
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
add a comment |
$begingroup$
Attempt:
Notice we can write $T = max (T_1, T_2)$. One can find the pmf:
$$ P(T leq t) = P(max(T_1,T_2) leq t) = P(T_1 leq t, T_2 leq t) = (1-e^{-t/2})^2$$
therefore $f_T(t) = (1-e^{t/2}) e^{-t/2} = e^{-t/2} - e^{-t}$
Now, $f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)} = dfrac{e^{-t/2} - e^{-t}}{e^{-1}} $
Thus, the expectation is
$$ E(T | T_1=1) = int t dfrac{e^{-t/2} - e^{-t}}{2e^{-1/2}} dt = e (2 - 1) = boxed{2e^{1/2}}$$
Now, for $b$, we evaluate $int_1^{infty} f_{T_1} (t) = 1 - P(1) = e^{-1/2} $
So, $E(T | T_1 > 1) = e^{1/2} $
Now, Is my approach correct so far? For part c), what would be the best approach? or just brute furce ?
probability-theory conditional-probability
edited Jan 27 at 10:34
Did
248k23226466
asked Jan 27 at 10:21
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
$begingroup$
Sorry but the formula $$f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)}$$ seems to be a pure invention. Note for example that $$int_0^inftyfrac{f_T(t)}{f_{T_1}(1)}dt=frac1{f_{T_1}(1)}ne1$$ which should ring as an alarm bell.
$endgroup$
– Did
Jan 27 at 10:33
$begingroup$
I think I misunderstood the probability mass function from the cumulative
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
$begingroup$
I think the correct formulation is $$ frac{ P(T leq t, T_1 leq t_1 ) }{P(T_1 leq t_1 ) } $$ and then differentiate and plug $t_1=1$
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
add a comment |
$begingroup$
Attempt:
Notice we can write $T = max (T_1, T_2)$. One can find the pmf:
$$ P(T leq t) = P(max(T_1,T_2) leq t) = P(T_1 leq t, T_2 leq t) = (1-e^{-t/2})^2$$
therefore $f_T(t) = (1-e^{t/2}) e^{-t/2} = e^{-t/2} - e^{-t}$
Now, $f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)} = dfrac{e^{-t/2} - e^{-t}}{e^{-1}} $
Thus, the expectation is
$$ E(T | T_1=1) = int t dfrac{e^{-t/2} - e^{-t}}{2e^{-1/2}} dt = e (2 - 1) = boxed{2e^{1/2}}$$
Now, for $b$, we evaluate $int_1^{infty} f_{T_1} (t) = 1 - P(1) = e^{-1/2} $
So, $E(T | T_1 > 1) = e^{1/2} $
Now, Is my approach correct so far? For part c), what would be the best approach? or just brute furce ?
probability-theory conditional-probability
edited Jan 27 at 10:34
Did
248k23226466
asked Jan 27 at 10:21
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
Attempt:
Notice we can write $T = max (T_1, T_2)$. One can find the pmf:
$$ P(T leq t) = P(max(T_1,T_2) leq t) = P(T_1 leq t, T_2 leq t) = (1-e^{-t/2})^2$$
therefore $f_T(t) = (1-e^{t/2}) e^{-t/2} = e^{-t/2} - e^{-t}$
Now, $f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)} = dfrac{e^{-t/2} - e^{-t}}{e^{-1}} $
Thus, the expectation is
$$ E(T | T_1=1) = int t dfrac{e^{-t/2} - e^{-t}}{2e^{-1/2}} dt = e (2 - 1) = boxed{2e^{1/2}}$$
Now, for $b$, we evaluate $int_1^{infty} f_{T_1} (t) = 1 - P(1) = e^{-1/2} $
So, $E(T | T_1 > 1) = e^{1/2} $
Now, Is my approach correct so far? For part c), what would be the best approach? or just brute furce ?
probability-theory conditional-probability
probability-theory conditional-probability
edited Jan 27 at 10:34
Did
248k23226466
asked Jan 27 at 10:21
Jimmy SabaterJimmy Sabater
2,539325
edited Jan 27 at 10:34
Did
248k23226466
asked Jan 27 at 10:21
Jimmy SabaterJimmy Sabater
2,539325
edited Jan 27 at 10:34
Did
248k23226466
edited Jan 27 at 10:34
Did
248k23226466
edited Jan 27 at 10:34
Did
248k23226466
248k23226466
asked Jan 27 at 10:21
Jimmy SabaterJimmy Sabater
2,539325
asked Jan 27 at 10:21
Jimmy SabaterJimmy Sabater
2,539325
asked Jan 27 at 10:21
Jimmy SabaterJimmy Sabater
2,539325
2,539325
$begingroup$
Sorry but the formula $$f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)}$$ seems to be a pure invention. Note for example that $$int_0^inftyfrac{f_T(t)}{f_{T_1}(1)}dt=frac1{f_{T_1}(1)}ne1$$ which should ring as an alarm bell.
$endgroup$
– Did
Jan 27 at 10:33
$begingroup$
I think I misunderstood the probability mass function from the cumulative
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
$begingroup$
I think the correct formulation is $$ frac{ P(T leq t, T_1 leq t_1 ) }{P(T_1 leq t_1 ) } $$ and then differentiate and plug $t_1=1$
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
add a comment |
$begingroup$
Sorry but the formula $$f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)}$$ seems to be a pure invention. Note for example that $$int_0^inftyfrac{f_T(t)}{f_{T_1}(1)}dt=frac1{f_{T_1}(1)}ne1$$ which should ring as an alarm bell.
$endgroup$
– Did
Jan 27 at 10:33
$begingroup$
I think I misunderstood the probability mass function from the cumulative
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
$begingroup$
I think the correct formulation is $$ frac{ P(T leq t, T_1 leq t_1 ) }{P(T_1 leq t_1 ) } $$ and then differentiate and plug $t_1=1$
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
$begingroup$
Sorry but the formula $$f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)}$$ seems to be a pure invention. Note for example that $$int_0^inftyfrac{f_T(t)}{f_{T_1}(1)}dt=frac1{f_{T_1}(1)}ne1$$ which should ring as an alarm bell.
$endgroup$
– Did
Jan 27 at 10:33
$begingroup$
Sorry but the formula $$f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)}$$ seems to be a pure invention. Note for example that $$int_0^inftyfrac{f_T(t)}{f_{T_1}(1)}dt=frac1{f_{T_1}(1)}ne1$$ which should ring as an alarm bell.
$endgroup$
– Did
Jan 27 at 10:33
$begingroup$
I think I misunderstood the probability mass function from the cumulative
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
$begingroup$
I think I misunderstood the probability mass function from the cumulative
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
$begingroup$
I think the correct formulation is $$ frac{ P(T leq t, T_1 leq t_1 ) }{P(T_1 leq t_1 ) } $$ and then differentiate and plug $t_1=1$
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
$begingroup$
I think the correct formulation is $$ frac{ P(T leq t, T_1 leq t_1 ) }{P(T_1 leq t_1 ) } $$ and then differentiate and plug $t_1=1$
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For a) I would consider
$$
E(T|T_1 = 1) = E(T|T_2 > 1, T_1 =1) P(T_2 > 1) + E(T|T_2<1, T_1=1)P(T_2<1)
$$
The first expected value simplifies to $E(T_2|T_2>1) = 1 +E(T_2)$ using memoryless property and the second one is just 1.
The same approach works for b)
$$
E(T|T_1 > 1) = E(T|T_2 > 1, T_1 > 1) P(T_2 > 1) + E(T|T_2<1, T_1> 1)P(T_2<1)
$$
Using memoryless property of $T_1$ and $T_2$ the first expected value is
$1 + E(T)$ and the second one is $E(T_1|T_1>1)=1+E(T_1)$.
Hope this helps for c) :)
answered Feb 2 at 11:07
RoahRoah
1,028917
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
For a) I would consider
$$
E(T|T_1 = 1) = E(T|T_2 > 1, T_1 =1) P(T_2 > 1) + E(T|T_2<1, T_1=1)P(T_2<1)
$$
The first expected value simplifies to $E(T_2|T_2>1) = 1 +E(T_2)$ using memoryless property and the second one is just 1.
The same approach works for b)
$$
E(T|T_1 > 1) = E(T|T_2 > 1, T_1 > 1) P(T_2 > 1) + E(T|T_2<1, T_1> 1)P(T_2<1)
$$
Using memoryless property of $T_1$ and $T_2$ the first expected value is
$1 + E(T)$ and the second one is $E(T_1|T_1>1)=1+E(T_1)$.
Hope this helps for c) :)
answered Feb 2 at 11:07
RoahRoah
1,028917
$endgroup$
add a comment |
$begingroup$
For a) I would consider
$$
E(T|T_1 = 1) = E(T|T_2 > 1, T_1 =1) P(T_2 > 1) + E(T|T_2<1, T_1=1)P(T_2<1)
$$
The first expected value simplifies to $E(T_2|T_2>1) = 1 +E(T_2)$ using memoryless property and the second one is just 1.
The same approach works for b)
$$
E(T|T_1 > 1) = E(T|T_2 > 1, T_1 > 1) P(T_2 > 1) + E(T|T_2<1, T_1> 1)P(T_2<1)
$$
Using memoryless property of $T_1$ and $T_2$ the first expected value is
$1 + E(T)$ and the second one is $E(T_1|T_1>1)=1+E(T_1)$.
Hope this helps for c) :)
answered Feb 2 at 11:07
RoahRoah
1,028917
$endgroup$
add a comment |
$begingroup$
For a) I would consider
$$
E(T|T_1 = 1) = E(T|T_2 > 1, T_1 =1) P(T_2 > 1) + E(T|T_2<1, T_1=1)P(T_2<1)
$$
The first expected value simplifies to $E(T_2|T_2>1) = 1 +E(T_2)$ using memoryless property and the second one is just 1.
The same approach works for b)
$$
E(T|T_1 > 1) = E(T|T_2 > 1, T_1 > 1) P(T_2 > 1) + E(T|T_2<1, T_1> 1)P(T_2<1)
$$
Using memoryless property of $T_1$ and $T_2$ the first expected value is
$1 + E(T)$ and the second one is $E(T_1|T_1>1)=1+E(T_1)$.
Hope this helps for c) :)
answered Feb 2 at 11:07
RoahRoah
1,028917
$endgroup$
For a) I would consider
$$
E(T|T_1 = 1) = E(T|T_2 > 1, T_1 =1) P(T_2 > 1) + E(T|T_2<1, T_1=1)P(T_2<1)
$$
The first expected value simplifies to $E(T_2|T_2>1) = 1 +E(T_2)$ using memoryless property and the second one is just 1.
The same approach works for b)
$$
E(T|T_1 > 1) = E(T|T_2 > 1, T_1 > 1) P(T_2 > 1) + E(T|T_2<1, T_1> 1)P(T_2<1)
$$
Using memoryless property of $T_1$ and $T_2$ the first expected value is
$1 + E(T)$ and the second one is $E(T_1|T_1>1)=1+E(T_1)$.
Hope this helps for c) :)
answered Feb 2 at 11:07
RoahRoah
1,028917
answered Feb 2 at 11:07
RoahRoah
1,028917
answered Feb 2 at 11:07
RoahRoah
1,028917
answered Feb 2 at 11:07
RoahRoah
1,028917
1,028917
add a comment |
add a comment |
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$begingroup$
Sorry but the formula $$f_{T|T_1=1} (t|t_1) = dfrac{f_T(t)}{f_{T_1}(1)}$$ seems to be a pure invention. Note for example that $$int_0^inftyfrac{f_T(t)}{f_{T_1}(1)}dt=frac1{f_{T_1}(1)}ne1$$ which should ring as an alarm bell.
$endgroup$
– Did
Jan 27 at 10:33
$begingroup$
I think I misunderstood the probability mass function from the cumulative
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34
$begingroup$
I think the correct formulation is $$ frac{ P(T leq t, T_1 leq t_1 ) }{P(T_1 leq t_1 ) } $$ and then differentiate and plug $t_1=1$
$endgroup$
– Jimmy Sabater
Jan 27 at 10:34