Mixing
Computing pmf and cdf for a function of an exponential random variable
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I'm a little stuck on this one due to the nature of the function. Here is the question:
$mathit{T}$ is a $lambda$ = 1 exponential random variable and $mathit{f(x)= lfloor xrfloor}$ (largest integer not more than $mathit{x}$).
Find the cdf and pmf of $mathit{X = f(T)}$. What is $mathbb{E}$[$mathit{f(T)}$]?
I don't know how to work with the "floor" portion of the question. I'm sure once I got past that I could do the rest of the question.
probability statistics probability-distributions exponential-distribution
$endgroup$
add a comment |
$begingroup$
I'm a little stuck on this one due to the nature of the function. Here is the question:
$mathit{T}$ is a $lambda$ = 1 exponential random variable and $mathit{f(x)= lfloor xrfloor}$ (largest integer not more than $mathit{x}$).
Find the cdf and pmf of $mathit{X = f(T)}$. What is $mathbb{E}$[$mathit{f(T)}$]?
I don't know how to work with the "floor" portion of the question. I'm sure once I got past that I could do the rest of the question.
probability statistics probability-distributions exponential-distribution
$endgroup$
$begingroup$
I suppose I'm really stuck on how to find X=f(T), because of the "floor" function on x
$endgroup$
– user610107
Jan 31 at 1:55
add a comment |
$begingroup$
I'm a little stuck on this one due to the nature of the function. Here is the question:
$mathit{T}$ is a $lambda$ = 1 exponential random variable and $mathit{f(x)= lfloor xrfloor}$ (largest integer not more than $mathit{x}$).
Find the cdf and pmf of $mathit{X = f(T)}$. What is $mathbb{E}$[$mathit{f(T)}$]?
I don't know how to work with the "floor" portion of the question. I'm sure once I got past that I could do the rest of the question.
probability statistics probability-distributions exponential-distribution
$endgroup$
I'm a little stuck on this one due to the nature of the function. Here is the question:
$mathit{T}$ is a $lambda$ = 1 exponential random variable and $mathit{f(x)= lfloor xrfloor}$ (largest integer not more than $mathit{x}$).
Find the cdf and pmf of $mathit{X = f(T)}$. What is $mathbb{E}$[$mathit{f(T)}$]?
I don't know how to work with the "floor" portion of the question. I'm sure once I got past that I could do the rest of the question.
probability statistics probability-distributions exponential-distribution
probability statistics probability-distributions exponential-distribution
asked Jan 31 at 1:47
user610107
$begingroup$
I suppose I'm really stuck on how to find X=f(T), because of the "floor" function on x
$endgroup$
– user610107
Jan 31 at 1:55
add a comment |
$begingroup$
I suppose I'm really stuck on how to find X=f(T), because of the "floor" function on x
$endgroup$
– user610107
Jan 31 at 1:55
$begingroup$
I suppose I'm really stuck on how to find X=f(T), because of the "floor" function on x
$endgroup$
– user610107
Jan 31 at 1:55
$begingroup$
I suppose I'm really stuck on how to find X=f(T), because of the "floor" function on x
$endgroup$
– user610107
Jan 31 at 1:55
add a comment |
1 Answer
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$begingroup$
Let $Y = lfloor X rfloor$, where $X sim operatorname{Exponential}(lambda)$ is exponentially distributed with rate $lambda$. Then clearly $Y in {0, 1, 2, ldots}$, since $X ge 0$ means $lfloor X rfloor$ takes on nonnegative integer values.
Specifically, $Y = 0$ if $0 le X < 1$, so $$Pr[Y = 0] = Pr[0 le X < 1] = F_X(1) - F_X(0) = 1 - e^{-lambda}.$$ Similarly, $Y = 1$ if $1 le X < 2$, so $$Pr[Y = 1] = Pr[1 le X < 2] = F_X(2) - F_X(1) = (1 - e^{-2lambda}) - (1 - e^{-lambda}) = e^{-lambda} - e^{-2lambda}.$$ And it is easy to see that in general, for a nonnegative integer $y$, $$Pr[Y = y] = Pr[y le X < y+1] = F_X(y+1) - F_X(y) = e^{-lambda y} - e^{-lambda(y+1)}.$$
The only part remaining is to simplify this expression and determine the kind of distribution $Y$ actually is.
answered Jan 31 at 2:15
heropupheropup
65.3k865104
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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active
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votes
$begingroup$
Let $Y = lfloor X rfloor$, where $X sim operatorname{Exponential}(lambda)$ is exponentially distributed with rate $lambda$. Then clearly $Y in {0, 1, 2, ldots}$, since $X ge 0$ means $lfloor X rfloor$ takes on nonnegative integer values.
Specifically, $Y = 0$ if $0 le X < 1$, so $$Pr[Y = 0] = Pr[0 le X < 1] = F_X(1) - F_X(0) = 1 - e^{-lambda}.$$ Similarly, $Y = 1$ if $1 le X < 2$, so $$Pr[Y = 1] = Pr[1 le X < 2] = F_X(2) - F_X(1) = (1 - e^{-2lambda}) - (1 - e^{-lambda}) = e^{-lambda} - e^{-2lambda}.$$ And it is easy to see that in general, for a nonnegative integer $y$, $$Pr[Y = y] = Pr[y le X < y+1] = F_X(y+1) - F_X(y) = e^{-lambda y} - e^{-lambda(y+1)}.$$
The only part remaining is to simplify this expression and determine the kind of distribution $Y$ actually is.
answered Jan 31 at 2:15
heropupheropup
65.3k865104
$endgroup$
add a comment |
$begingroup$
Let $Y = lfloor X rfloor$, where $X sim operatorname{Exponential}(lambda)$ is exponentially distributed with rate $lambda$. Then clearly $Y in {0, 1, 2, ldots}$, since $X ge 0$ means $lfloor X rfloor$ takes on nonnegative integer values.
Specifically, $Y = 0$ if $0 le X < 1$, so $$Pr[Y = 0] = Pr[0 le X < 1] = F_X(1) - F_X(0) = 1 - e^{-lambda}.$$ Similarly, $Y = 1$ if $1 le X < 2$, so $$Pr[Y = 1] = Pr[1 le X < 2] = F_X(2) - F_X(1) = (1 - e^{-2lambda}) - (1 - e^{-lambda}) = e^{-lambda} - e^{-2lambda}.$$ And it is easy to see that in general, for a nonnegative integer $y$, $$Pr[Y = y] = Pr[y le X < y+1] = F_X(y+1) - F_X(y) = e^{-lambda y} - e^{-lambda(y+1)}.$$
The only part remaining is to simplify this expression and determine the kind of distribution $Y$ actually is.
answered Jan 31 at 2:15
heropupheropup
65.3k865104
$endgroup$
add a comment |
$begingroup$
Let $Y = lfloor X rfloor$, where $X sim operatorname{Exponential}(lambda)$ is exponentially distributed with rate $lambda$. Then clearly $Y in {0, 1, 2, ldots}$, since $X ge 0$ means $lfloor X rfloor$ takes on nonnegative integer values.
Specifically, $Y = 0$ if $0 le X < 1$, so $$Pr[Y = 0] = Pr[0 le X < 1] = F_X(1) - F_X(0) = 1 - e^{-lambda}.$$ Similarly, $Y = 1$ if $1 le X < 2$, so $$Pr[Y = 1] = Pr[1 le X < 2] = F_X(2) - F_X(1) = (1 - e^{-2lambda}) - (1 - e^{-lambda}) = e^{-lambda} - e^{-2lambda}.$$ And it is easy to see that in general, for a nonnegative integer $y$, $$Pr[Y = y] = Pr[y le X < y+1] = F_X(y+1) - F_X(y) = e^{-lambda y} - e^{-lambda(y+1)}.$$
The only part remaining is to simplify this expression and determine the kind of distribution $Y$ actually is.
answered Jan 31 at 2:15
heropupheropup
65.3k865104
$endgroup$
Let $Y = lfloor X rfloor$, where $X sim operatorname{Exponential}(lambda)$ is exponentially distributed with rate $lambda$. Then clearly $Y in {0, 1, 2, ldots}$, since $X ge 0$ means $lfloor X rfloor$ takes on nonnegative integer values.
Specifically, $Y = 0$ if $0 le X < 1$, so $$Pr[Y = 0] = Pr[0 le X < 1] = F_X(1) - F_X(0) = 1 - e^{-lambda}.$$ Similarly, $Y = 1$ if $1 le X < 2$, so $$Pr[Y = 1] = Pr[1 le X < 2] = F_X(2) - F_X(1) = (1 - e^{-2lambda}) - (1 - e^{-lambda}) = e^{-lambda} - e^{-2lambda}.$$ And it is easy to see that in general, for a nonnegative integer $y$, $$Pr[Y = y] = Pr[y le X < y+1] = F_X(y+1) - F_X(y) = e^{-lambda y} - e^{-lambda(y+1)}.$$
The only part remaining is to simplify this expression and determine the kind of distribution $Y$ actually is.
answered Jan 31 at 2:15
heropupheropup
65.3k865104
answered Jan 31 at 2:15
heropupheropup
65.3k865104
answered Jan 31 at 2:15
heropupheropup
65.3k865104
answered Jan 31 at 2:15
heropupheropup
65.3k865104
65.3k865104
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$begingroup$
I suppose I'm really stuck on how to find X=f(T), because of the "floor" function on x
$endgroup$
– user610107
Jan 31 at 1:55