Calculating the expectation of types
$begingroup$
A certain region is inhabited by $r$ distinct types of a species of insect. Each insect caught will, independently of the types of the previous catches, be of the type $i$ with probability
$$P_i, i= 1,dots,r$$. Calculate the mean number of types of insects that are caught before the first type 1 catch.
The solution I am looking at defines an indicator as
$$X_j = 1$$ when type j is caught before type 1
$$X_j = 0$$ otherwise
Then $$mathbb{E}[X_j] = P{ textrm{type j before type 1}} = P{ j|textrm{j or i}} = frac{P_j}{P_j + P_1}$$
But I don’t understand how
$$P{ textrm{type j before type 1}} = P{ j|textrm{j or i}} $$
Can someone explain how this step is correct?
probability probability-distributions random-variables expected-value
asked Dec 26 '18 at 20:46
user601297user601297
41719
$endgroup$
add a comment |
$begingroup$
A certain region is inhabited by $r$ distinct types of a species of insect. Each insect caught will, independently of the types of the previous catches, be of the type $i$ with probability
$$P_i, i= 1,dots,r$$. Calculate the mean number of types of insects that are caught before the first type 1 catch.
The solution I am looking at defines an indicator as
$$X_j = 1$$ when type j is caught before type 1
$$X_j = 0$$ otherwise
Then $$mathbb{E}[X_j] = P{ textrm{type j before type 1}} = P{ j|textrm{j or i}} = frac{P_j}{P_j + P_1}$$
But I don’t understand how
$$P{ textrm{type j before type 1}} = P{ j|textrm{j or i}} $$
Can someone explain how this step is correct?
probability probability-distributions random-variables expected-value
asked Dec 26 '18 at 20:46
user601297user601297
41719
$endgroup$
add a comment |
$begingroup$
A certain region is inhabited by $r$ distinct types of a species of insect. Each insect caught will, independently of the types of the previous catches, be of the type $i$ with probability
$$P_i, i= 1,dots,r$$. Calculate the mean number of types of insects that are caught before the first type 1 catch.
The solution I am looking at defines an indicator as
$$X_j = 1$$ when type j is caught before type 1
$$X_j = 0$$ otherwise
Then $$mathbb{E}[X_j] = P{ textrm{type j before type 1}} = P{ j|textrm{j or i}} = frac{P_j}{P_j + P_1}$$
But I don’t understand how
$$P{ textrm{type j before type 1}} = P{ j|textrm{j or i}} $$
Can someone explain how this step is correct?
probability probability-distributions random-variables expected-value
asked Dec 26 '18 at 20:46
user601297user601297
41719
$endgroup$
A certain region is inhabited by $r$ distinct types of a species of insect. Each insect caught will, independently of the types of the previous catches, be of the type $i$ with probability
$$P_i, i= 1,dots,r$$. Calculate the mean number of types of insects that are caught before the first type 1 catch.
The solution I am looking at defines an indicator as
$$X_j = 1$$ when type j is caught before type 1
$$X_j = 0$$ otherwise
Then $$mathbb{E}[X_j] = P{ textrm{type j before type 1}} = P{ j|textrm{j or i}} = frac{P_j}{P_j + P_1}$$
But I don’t understand how
$$P{ textrm{type j before type 1}} = P{ j|textrm{j or i}} $$
Can someone explain how this step is correct?
probability probability-distributions random-variables expected-value
probability probability-distributions random-variables expected-value
asked Dec 26 '18 at 20:46
user601297user601297
41719
asked Dec 26 '18 at 20:46
user601297user601297
41719
asked Dec 26 '18 at 20:46
user601297user601297
41719
asked Dec 26 '18 at 20:46
user601297user601297
41719
asked Dec 26 '18 at 20:46
user601297user601297
41719
41719
add a comment |
add a comment |
1 Answer
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$begingroup$
When we think about the event $X_j$, we are only interested in which is caught first: an insect of type $1$ or an insect of type $j$. And to determine which is caught first, we only need to think about the first time that a type $1$ or a type $j$ is caught (call this event $C_j$). This is because once $C_j$ happens then $X_j$ is determined, and any captures before $C_j$ don't affect $X_j$.
So at the capture $C_j$ we know that either a type $j$ or a type $1$ was captured. At $C_j$, the probability that a type $j$ was captured is
$$mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{mathbb{P}(textrm{type } j textrm{ captured} cap (textrm{type } j textrm{ or } 1textrm{ captured}))}{mathbb{P}(textrm{type } j textrm{ or } 1textrm{ captured})}.$$
The numerator is just the probability that an insect of type $j$ is captured, so has probability $P_j$. The denominator has probability $P_1 + P_j$. Substituting this into the formula above, the probability of $X_j$ is the probability that a type $j$ was chosen at capture $C_j$, which is
$$mathbb{P}(X_j) = mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{P_j}{P_1 + P_j}.$$
answered Dec 27 '18 at 11:35
AlexAlex
744412
$endgroup$
add a comment |
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1 Answer
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$begingroup$
When we think about the event $X_j$, we are only interested in which is caught first: an insect of type $1$ or an insect of type $j$. And to determine which is caught first, we only need to think about the first time that a type $1$ or a type $j$ is caught (call this event $C_j$). This is because once $C_j$ happens then $X_j$ is determined, and any captures before $C_j$ don't affect $X_j$.
So at the capture $C_j$ we know that either a type $j$ or a type $1$ was captured. At $C_j$, the probability that a type $j$ was captured is
$$mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{mathbb{P}(textrm{type } j textrm{ captured} cap (textrm{type } j textrm{ or } 1textrm{ captured}))}{mathbb{P}(textrm{type } j textrm{ or } 1textrm{ captured})}.$$
The numerator is just the probability that an insect of type $j$ is captured, so has probability $P_j$. The denominator has probability $P_1 + P_j$. Substituting this into the formula above, the probability of $X_j$ is the probability that a type $j$ was chosen at capture $C_j$, which is
$$mathbb{P}(X_j) = mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{P_j}{P_1 + P_j}.$$
answered Dec 27 '18 at 11:35
AlexAlex
744412
$endgroup$
add a comment |
$begingroup$
When we think about the event $X_j$, we are only interested in which is caught first: an insect of type $1$ or an insect of type $j$. And to determine which is caught first, we only need to think about the first time that a type $1$ or a type $j$ is caught (call this event $C_j$). This is because once $C_j$ happens then $X_j$ is determined, and any captures before $C_j$ don't affect $X_j$.
So at the capture $C_j$ we know that either a type $j$ or a type $1$ was captured. At $C_j$, the probability that a type $j$ was captured is
$$mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{mathbb{P}(textrm{type } j textrm{ captured} cap (textrm{type } j textrm{ or } 1textrm{ captured}))}{mathbb{P}(textrm{type } j textrm{ or } 1textrm{ captured})}.$$
The numerator is just the probability that an insect of type $j$ is captured, so has probability $P_j$. The denominator has probability $P_1 + P_j$. Substituting this into the formula above, the probability of $X_j$ is the probability that a type $j$ was chosen at capture $C_j$, which is
$$mathbb{P}(X_j) = mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{P_j}{P_1 + P_j}.$$
answered Dec 27 '18 at 11:35
AlexAlex
744412
$endgroup$
add a comment |
$begingroup$
When we think about the event $X_j$, we are only interested in which is caught first: an insect of type $1$ or an insect of type $j$. And to determine which is caught first, we only need to think about the first time that a type $1$ or a type $j$ is caught (call this event $C_j$). This is because once $C_j$ happens then $X_j$ is determined, and any captures before $C_j$ don't affect $X_j$.
So at the capture $C_j$ we know that either a type $j$ or a type $1$ was captured. At $C_j$, the probability that a type $j$ was captured is
$$mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{mathbb{P}(textrm{type } j textrm{ captured} cap (textrm{type } j textrm{ or } 1textrm{ captured}))}{mathbb{P}(textrm{type } j textrm{ or } 1textrm{ captured})}.$$
The numerator is just the probability that an insect of type $j$ is captured, so has probability $P_j$. The denominator has probability $P_1 + P_j$. Substituting this into the formula above, the probability of $X_j$ is the probability that a type $j$ was chosen at capture $C_j$, which is
$$mathbb{P}(X_j) = mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{P_j}{P_1 + P_j}.$$
answered Dec 27 '18 at 11:35
AlexAlex
744412
$endgroup$
When we think about the event $X_j$, we are only interested in which is caught first: an insect of type $1$ or an insect of type $j$. And to determine which is caught first, we only need to think about the first time that a type $1$ or a type $j$ is caught (call this event $C_j$). This is because once $C_j$ happens then $X_j$ is determined, and any captures before $C_j$ don't affect $X_j$.
So at the capture $C_j$ we know that either a type $j$ or a type $1$ was captured. At $C_j$, the probability that a type $j$ was captured is
$$mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{mathbb{P}(textrm{type } j textrm{ captured} cap (textrm{type } j textrm{ or } 1textrm{ captured}))}{mathbb{P}(textrm{type } j textrm{ or } 1textrm{ captured})}.$$
The numerator is just the probability that an insect of type $j$ is captured, so has probability $P_j$. The denominator has probability $P_1 + P_j$. Substituting this into the formula above, the probability of $X_j$ is the probability that a type $j$ was chosen at capture $C_j$, which is
$$mathbb{P}(X_j) = mathbb{P}(textrm{type } j textrm{ captured} mid textrm{type } j textrm{ or } 1textrm{ captured}) = frac{P_j}{P_1 + P_j}.$$
answered Dec 27 '18 at 11:35
AlexAlex
744412
edited Dec 27 '18 at 11:40
answered Dec 27 '18 at 11:35
AlexAlex
744412
answered Dec 27 '18 at 11:35
AlexAlex
744412
answered Dec 27 '18 at 11:35
AlexAlex
744412
744412
add a comment |
add a comment |
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