Sum of independent random variables with different distributions
Can we find the distribution of the sum of random variables with different pmf and different possible values?For example let X be a Poisson random variable with parameter $lambda$ and Y an independent Bernoulli random variable with parameter $p$. Then is the probability mass function of X + Y
$$p_{X+Y}(n)=sum_{k=1}^n P(X=n-k)P(Y=k) = e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p = pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}?$$
probability proof-verification
asked Nov 22 at 14:27
dxdydz
1949
add a comment |
Can we find the distribution of the sum of random variables with different pmf and different possible values?For example let X be a Poisson random variable with parameter $lambda$ and Y an independent Bernoulli random variable with parameter $p$. Then is the probability mass function of X + Y
$$p_{X+Y}(n)=sum_{k=1}^n P(X=n-k)P(Y=k) = e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p = pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}?$$
probability proof-verification
asked Nov 22 at 14:27
dxdydz
1949
add a comment |
Can we find the distribution of the sum of random variables with different pmf and different possible values?For example let X be a Poisson random variable with parameter $lambda$ and Y an independent Bernoulli random variable with parameter $p$. Then is the probability mass function of X + Y
$$p_{X+Y}(n)=sum_{k=1}^n P(X=n-k)P(Y=k) = e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p = pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}?$$
probability proof-verification
asked Nov 22 at 14:27
dxdydz
1949
Can we find the distribution of the sum of random variables with different pmf and different possible values?For example let X be a Poisson random variable with parameter $lambda$ and Y an independent Bernoulli random variable with parameter $p$. Then is the probability mass function of X + Y
$$p_{X+Y}(n)=sum_{k=1}^n P(X=n-k)P(Y=k) = e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p = pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}?$$
probability proof-verification
probability proof-verification
asked Nov 22 at 14:27
dxdydz
1949
asked Nov 22 at 14:27
dxdydz
1949
asked Nov 22 at 14:27
dxdydz
1949
asked Nov 22 at 14:27
dxdydz
1949
asked Nov 22 at 14:27
dxdydz
1949
1949
add a comment |
add a comment |
1 Answer
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if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$
if $n>0$, then we have
begin{align}
p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
&= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
end{align}
answered Nov 22 at 14:34
Siong Thye Goh
98.4k1463116
add a comment |
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1 Answer
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1 Answer
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if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$
if $n>0$, then we have
begin{align}
p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
&= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
end{align}
answered Nov 22 at 14:34
Siong Thye Goh
98.4k1463116
add a comment |
if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$
if $n>0$, then we have
begin{align}
p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
&= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
end{align}
answered Nov 22 at 14:34
Siong Thye Goh
98.4k1463116
add a comment |
if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$
if $n>0$, then we have
begin{align}
p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
&= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
end{align}
answered Nov 22 at 14:34
Siong Thye Goh
98.4k1463116
if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$
if $n>0$, then we have
begin{align}
p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
&= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
end{align}
answered Nov 22 at 14:34
Siong Thye Goh
98.4k1463116
answered Nov 22 at 14:34
Siong Thye Goh
98.4k1463116
answered Nov 22 at 14:34
Siong Thye Goh
98.4k1463116
answered Nov 22 at 14:34
Siong Thye Goh
98.4k1463116
98.4k1463116
add a comment |
add a comment |
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