Suppose I have two discrete random variables U and S, which are independent. I want to find the expected...
$begingroup$
Suppose I have two independent discrete random variables U and S, with the following density functions:
$$f_U(k) = [1-(1-p)^k]^2 - [1-(1-p)^{k-1}]^2$$
$$f_S(j) = (1-(1-p)^2)[(1-p)^2]^{j-1} $$
That is, S is geometrical distributed with parameter $1 - (1-p)^2$.
I want to find $mathbb{E}[U+S]$. I tried using the definition of expected value of a discrete random variable: $mathbb{E}[X] = sum_{i=1}^infty x_i p_i$.
However the calculations have gotten a little out of hand. At least it seems to me. Do you have any idea how to approach this?
probability
asked Dec 5 '18 at 17:04
qcc101qcc101
597213
$endgroup$
add a comment |
$begingroup$
Suppose I have two independent discrete random variables U and S, with the following density functions:
$$f_U(k) = [1-(1-p)^k]^2 - [1-(1-p)^{k-1}]^2$$
$$f_S(j) = (1-(1-p)^2)[(1-p)^2]^{j-1} $$
That is, S is geometrical distributed with parameter $1 - (1-p)^2$.
I want to find $mathbb{E}[U+S]$. I tried using the definition of expected value of a discrete random variable: $mathbb{E}[X] = sum_{i=1}^infty x_i p_i$.
However the calculations have gotten a little out of hand. At least it seems to me. Do you have any idea how to approach this?
probability
asked Dec 5 '18 at 17:04
qcc101qcc101
597213
$endgroup$
4
$begingroup$
One knows that $E(U+S)=E(U)+E(S)$ as soon as $E(U)$ and $E(S)$ exist, for any random variables $U$ and $S$, independent or not. Source: Any first probability course.
$endgroup$
– Did
Dec 5 '18 at 17:08
add a comment |
$begingroup$
Suppose I have two independent discrete random variables U and S, with the following density functions:
$$f_U(k) = [1-(1-p)^k]^2 - [1-(1-p)^{k-1}]^2$$
$$f_S(j) = (1-(1-p)^2)[(1-p)^2]^{j-1} $$
That is, S is geometrical distributed with parameter $1 - (1-p)^2$.
I want to find $mathbb{E}[U+S]$. I tried using the definition of expected value of a discrete random variable: $mathbb{E}[X] = sum_{i=1}^infty x_i p_i$.
However the calculations have gotten a little out of hand. At least it seems to me. Do you have any idea how to approach this?
probability
asked Dec 5 '18 at 17:04
qcc101qcc101
597213
$endgroup$
Suppose I have two independent discrete random variables U and S, with the following density functions:
$$f_U(k) = [1-(1-p)^k]^2 - [1-(1-p)^{k-1}]^2$$
$$f_S(j) = (1-(1-p)^2)[(1-p)^2]^{j-1} $$
That is, S is geometrical distributed with parameter $1 - (1-p)^2$.
I want to find $mathbb{E}[U+S]$. I tried using the definition of expected value of a discrete random variable: $mathbb{E}[X] = sum_{i=1}^infty x_i p_i$.
However the calculations have gotten a little out of hand. At least it seems to me. Do you have any idea how to approach this?
probability
probability
asked Dec 5 '18 at 17:04
qcc101qcc101
597213
asked Dec 5 '18 at 17:04
qcc101qcc101
597213
asked Dec 5 '18 at 17:04
qcc101qcc101
597213
asked Dec 5 '18 at 17:04
qcc101qcc101
597213
asked Dec 5 '18 at 17:04
qcc101qcc101
597213
597213
4
$begingroup$
One knows that $E(U+S)=E(U)+E(S)$ as soon as $E(U)$ and $E(S)$ exist, for any random variables $U$ and $S$, independent or not. Source: Any first probability course.
$endgroup$
– Did
Dec 5 '18 at 17:08
add a comment |
4
$begingroup$
One knows that $E(U+S)=E(U)+E(S)$ as soon as $E(U)$ and $E(S)$ exist, for any random variables $U$ and $S$, independent or not. Source: Any first probability course.
$endgroup$
– Did
Dec 5 '18 at 17:08
4
4
$begingroup$
One knows that $E(U+S)=E(U)+E(S)$ as soon as $E(U)$ and $E(S)$ exist, for any random variables $U$ and $S$, independent or not. Source: Any first probability course.
$endgroup$
– Did
Dec 5 '18 at 17:08
$begingroup$
One knows that $E(U+S)=E(U)+E(S)$ as soon as $E(U)$ and $E(S)$ exist, for any random variables $U$ and $S$, independent or not. Source: Any first probability course.
$endgroup$
– Did
Dec 5 '18 at 17:08
add a comment |
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$begingroup$
One knows that $E(U+S)=E(U)+E(S)$ as soon as $E(U)$ and $E(S)$ exist, for any random variables $U$ and $S$, independent or not. Source: Any first probability course.
$endgroup$
– Did
Dec 5 '18 at 17:08