- #1

EngWiPy

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My main problem is to find the cumulative distribution function (CDF) of the following random variable (RV)

[tex]Y=\sum_{k=1}^K\underbrace{\frac{a_k}{b_k}}_{X_k}[/tex]

where ##\{a_k, b_k\}## are independent and identically distributed (i.i.d.) exponential RVs with parameter 1.

At first, I wanted to use the moment generating function (MGF), but some pointed out that the RVs ##X_k## have undefined means, and the MGF doesn't work in this case. They suggested the characteristic function (CF), because the CF always exists even if the means don't.

My fist step was then to find the CF of the RV ##Y## as follows

[tex]\psi_Y(t)=\text{E}_Y\{e^{jtx}\}=\text{E}_{X_1,\,X_2,\ldots,\,X_K}\{e^{jt\sum_{k=1}^KX_k}\}=\prod_{k=1}^K\text{E}_{X_k}\{e^{jtx_k}\}=\prod_{k=1}^K\psi_{X_k}(t)[/tex]

So, the next step would be to find ##\psi_{X_k}(t)##, which is given by

[tex]\psi_{X_k}(t) = \text{E}_{X_k}\{e^{jtx_k}\} = \int_0^{\infty}e^{jtx_k}\,f_{X_k}(x_k)\,dx_k[/tex]

where ##f_{X_k}(x_k)## is the probability distribution function (PDF) of the RV ##X_k##. It can be shown that ##f_{X_k}(x_k)=1/(1+x_k)^2##, and thus

[tex]\psi_{X_k}(t) = \int_0^{\infty}\frac{e^{jtx_k}}{(1+x_k)^2}\,dx_k[/tex]

My first question is how to evaluate this integral?

My approach: by integration by parts, this results in

[tex]\psi_{X_k}(t) = 1+jt\int_0^{\infty}\frac{e^{jtx_k}}{1+x_k}\,dx_k=1+jt\int_0^{\infty}\frac{\cos(tx_k)}{1+x_k}\,dx_k-t\int_0^{\infty}\frac{\sin(tx_k)}{1+x_k}\,dx_k[/tex]

which can be written in terms of cosine and sine integrals as

[tex]\psi_{X_k}(t) = 1+jt\left[-\sin(t)\text{si}(t)-\cos(t)\text{ci}(t)\right]-t\left[\sin(t)\text{ci}(t)-\cos(t)\text{si}(t)\right][/tex]

where ##\text{si}(t)=-\int_t^{\infty}\sin(z)/z\,dz## and ##\text{ci}(t)=-\int_t^{\infty}\cos(z)/z\,dz##.

Is what I did so far correct? I appreciate if some comment on this, because apparently I made a mistake in my mathematical analysis that resulted in an indefinite integral, but not sure where my mistake is.

Thanks in advance