1.2 Sums of Independent Random Variables
Let \(X\) and \(Y\) be independent random variables and \(Z=X+Y\), where \(X\) and \(Y\) are continuous random variables. \begin {align*} F_Z(z) &=P(Z\leq z)=P(X+Y\leq z)\\\\ & =\iint \limits _{x+y\leq z}f_{X,Y}(x,y)\,dx\,dy =\iint \limits _{x+y\leq z}f_X(x)f_Y(y)\,dx\,dy\\\\ &=\int \limits ^{\infty }_{-\infty }\int \limits ^{z-y}_{-\infty }f_X(x)\,f_Y(y)\,dx\,dy\\ \end {align*}
\[\implies \hspace {0.4cm} F_{X+Y}(z)=\int ^{\infty }_{-\infty }F_X(z-y)f_Y(y)dy\dots .......1.2.1\]
Differentiating 1.2.1 with respect to \(z\)
\[f_Z(z)=\int ^{\infty }_{-\infty }f_X(z-y)\,f_Y(y)\,dy\dots ......1.2.2.\]
Example 1.2.1. Let \(X\) and \(Y\) be independent Uniformly distributed random variables on \((0,2)\), find
the pdf of \(X+Y\)
\[f_X(x)=f_Y(y)=\frac {1}{2},\hspace {0.5cm} x,y\in (0,2)\]
Then \(\hspace {0.2cm}\displaystyle { f_Z(z)=\int ^z_0\frac {1}{2}\,\cdot \,\frac {1}{2}\,dy=\frac {z}{4}\hspace {0.3cm}\text {if}\hspace {0.3cm} 0<z<2}\)
\[f_Z(z)=\int _{z-2}^2\frac {1}{4}\,dy=\frac {y}{4}\Big |^2_{z-2}=1-\frac {z}{4}\hspace {0.4cm}\text {if}\hspace {0.3cm} 2<z<4\]
\[ f_Z(z)= \begin {cases} \dfrac {z}{4},& 0<z<2\\\\ 1-\dfrac {z}{4},&2<z<4\\ \end {cases} \]
Distribution of the Sum of two Independent Poisson Distributions
Let \(X\) and \(Y\) be independent Poisson random variables with mean \(\mu _1\) and \(\mu _2\) respectively. Find the
distribution of \(Z=X+Y\)
\begin {align*} P\Big (Z=z\Big ) &=P\Big (X+Y=z\Big ) =\sum ^z_{x=0}P\Big (X=x, Y=z-x\Big )\\ &=\sum ^z_{x=0}P(X=x)P(Y=z-x)\\ &=\sum ^z_{x=0}\frac {\mu _1^xe^{-\mu _1}}{x!}\hspace {0.2cm}\frac {\mu _2^{z-x}e^{-\mu }}{(z-x)!}\\ &=e^{-(\mu _1+\mu _2)}\sum ^z_{x=0}\frac {\mu _1^x\mu _2^{z-x}}{x!(z-x)!}\\ &=\frac {e^{-(\mu _1+\mu _2)}}{z!}\sum ^z_{x=0} \begin {pmatrix} z\\x\\ \end {pmatrix} \mu _1^x\mu ^{z-x}_2\\ &=\frac {e^{-(\mu _1+\mu _2)}}{z!}(\mu _1+\mu _2)^z\\\\ &=\frac {(\mu _1+\mu _2)^ze^{-(\mu _1+\mu _2)}}{z!}\\ \end {align*}
\[\therefore \hspace {0.4cm} P(Z=z)=\frac {(\mu _1+\mu _2)^ze^{-(\mu _1+\mu _2)}}{z!},\hspace {0.5cm} Z\thicksim P(\mu _1+\mu _2)\]
* If \(X\) and \(Y\) are independent, then
\[M_Z(t)=E\Big (e^{tZ}\Big )=E\Big (e^{(Xt+Yt)}\Big )=E\Big (e^{Xt}\Big )E\Big (e^{Yt}\Big )=M_X(t)M_Y(t).\]
. If \(X_1,X_2,\dots ,X_n\) are independent Poisson such that \(E(X_j)=\mu _j\). Then \(Z\thicksim \sum \limits ^n_{j=1}\thicksim P\Big (\sum \limits ^n_{j=1}\Big )\)
. Prove by mathematical induction.
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