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In this problem, I am told that $X_1,X_2,X_3$ are mutually independent Poisson random variables with means $2,1,4$ respectively. I am also told to find the moment generating function for $Y=X_1+X_2+X_3$, which I found to be $M(t)=e^{7(e^t-1)}$. My question is how do I find $P(Y=5)$?. Is there a connection between $P(Y=5)$ and $M(t)$?

Attempt at a solution:

Since $X_1,X_2,X_3$ are mutually indpt. Poisson R.V. with mean $2,1,4$, we know that $$f(X_1)=\frac{e^{-2}2^x}{x!},$$ $$f(X_2)=\frac{e^{-1}1^x}{x!}$$ and $$f(X_3)=\frac{e^{-4}4^x}{x!}$$, which imply that $f(Y)=f(X_1)f(X_2)f(X_3)$ (is this true?I think it is bc of independence) Hence, after computations $$f(Y)=\frac{e^{-7}2^{3x}}{x!x!x!}.$$ To find $P(Y=5)$, do I just plug in $x=5$ into $f(Y)$? or do I take the summation of $f(Y)$ for $x=0$ all the way up to $x=5$ or neither? It's been a while since I have seen this material. Any help will be appreciated. Thanks!

Michael Hardy
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An P.
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2 Answers2

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The moment generating function of $Y$ is the mgf of a Poisson parameter $7$, so $Y$ has Poisson distribution parameter $7$. The required probability is $e^{-7}\frac{7^5}{5!}$.

Remark: We can prove by a direct computation of probabilities that if $U$ and $V$ are independent Poisson with parameters $\lambda$ and $\mu$, then $U+V$ has Poisson distribution with parameter $\lambda+\mu$. The result is also almost obvious if one considers the defining properties of the Poisson. However, the mgf approach is quick, and is a useful exercise.

André Nicolas
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$f(Y)=f(X_1 )f(X_2 )f(X_3)$ (is this true? I think it is bc of independence)

No.   Not in the slightest.   Independence means that the joint probability mass of the three random variables is the product of their probability masses. $$f_{X_1, X_2, X_3}(x_1, x_2, x_3) = f_{X_1}(x_1)\cdot f_{X_2}(x_2)\cdot f_{X_3}(x_3)$$

That is quite different from the probability mass of their sum, which is rather convoluted.

$$f_Y(y) = \sum_{x_1=0}^y\sum_{x_2=0}^{y-x_1} f_{X_1}(x_1)f_{X_2}(x_2)f_{X_3}(y-x_1-x_2)$$

Which can be solved but is lengthy and using moment generating function to handle the load is much easier.   The moment generating function of a sum of independent random variables is the product of their moment generating functions.   $$M_Y(t) = M_{X_1}(t)\cdot M_{X_2}(t)\cdot M_{X_3}(t)$$

Hence you found that $M_Y(t)=\mathsf e^{7\mathsf (e^t-1)}$.   Which you should recognise as the moment generation function of a Poisson Distributed random variable with rate parameter $7$.   Thus:

$$f_Y(y) ~=~ \dfrac{7^y\,\mathsf e^{-7}}{y!}\;\mathbf 1_{y\in \Bbb N}$$

Graham Kemp
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