We have a summation of independent discrete random variables (rvs) $Y = X_1 + X_2 + \ldots + X_n$. Assume the rvs can take non-negative real values. How can we find the probability mass function of $Y$? Is there any efficient method like the convolution for integer case?
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In the title you call these discrete random variables and refer to the mass function (consistent with discrete rvs). Do you mean they are supported on a countable subset of $\mathbb R^+$? If so, you might give a concrete example to clarify your intent, as the question is being interpreted otherwise. – Erick Wong May 09 '13 at 15:20
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Thank you. Yes, I mean discrete rvs. As a simple example that can be done by hand, let $X_1 \in {\omega_1 . (0, 1, 2)}$ and $X_2 \in {\omega_2 . (0, 1, 2)}$ such that $\omega_i \in \mathbb{R^+}$. There is no restriction on the pmfs $p(x_1)$ and $p(x_2)$, i.e, they can be any functions. Now, I am looking for a method to find the pmf of $Y = X_1 + X_2$. – Mas May 09 '13 at 16:32
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Use MGF: $$ tY=t X_1 +t X_2 +\ldots t + X_n\\ e^{tY}=e^{t(X_1 +X_2 +\ldots + X_n)}\\ \varphi_Y(t)=\varphi_{X_1}(t) \cdot \varphi_{X_2}(t) \cdot \ldots \cdot \varphi_{X_n}(t) $$ if, in addition, your $X_k$ are identically distributed, you get $$ \varphi_Y(t)=\varphi^{n}_{X_1}(t) $$
Alex
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Since the random variables are continuous, you would speak of their probability density function (instead of the probability mass function). The probability density function (PDF) of $Y$ is simply the (continuous) convolution of the PDFs of the random variables $X_i$. Convolution of two continuous random variables is defined by
$$(p_{X_1}*p_{X_2})(x)=\int_{-\infty}^{\infty}p_{X_1}(x-y)p_{X_2}(y)\;dy$$
EDIT: I was assuming your RVs are continuous, but maybe I misunderstood the question. Anyway, if they are discrete then (discrete) convolution is also the correct answer.
EXAMPLE: Let $X_1$ and $X_2$ be two discrete random variables, where $X_1$ takes on values $1/2$ and $3/4$ with probabilities $0.5$, and $X_2$ takes on values $1/8$ and $1/4$ with probabilities $0.4$ and $0.6$, respectively. So we have $p_{X_1}(x)=0.5\delta(x-1/2) + 0.5\delta(x-3/4)$ and $p_{X_2}=0.4\delta(x-1/8)+0.6\delta(x-1/4)$. Let $Y=X_1+X_2$. Then $p_Y(x)$ is given by the convolution of $p_{X_1}(x)$ and $p_{X_2}(x)$:
$$p_Y(x)=0.2\delta(x-5/8)+0.3\delta(x-3/4)+0.2\delta(x-7/8)+0.3\delta(x-1)$$
Matt L.
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Thank you for your reply. What you have done is totally right. You have converted the problem from a discrete case to a continuous one and then used the continuous convolution to solve it. But I am still not able to write a programming code to implement this cont convolution. I know that, if we have continuous signals, we can sample them with high rate and apply the discrete convolution to get an approximation to the continuous one. But here we have delta functions and we may miss some of them during the sampling process. – Mas May 09 '13 at 20:44
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What you have to do is make a list of all possible combinations of $X_1+X_2$ and weigh them by multiplying the corresponding probabilities of the individual outcomes, just like in the example above. You don't need to think in terms of continuous convolution. – Matt L. May 09 '13 at 21:57
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I was thinking for a convolution-like method to sum the output probabilities of the same outcome in one step. I think that I have to make it simple and solve it step by step. Thanks for your opinion. – Mas May 09 '13 at 22:44
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@Mas It's likely that a straightforward dynamic programming solution is best, unless the values have a very specific structure (for instance if they form a generalized arithmetic progression, then a higher-dimensional convolution should work). – Erick Wong May 10 '13 at 05:04